State the laws of electrostatics. Two electric charges of +15µC and +1µC are spaced 50 cm apart in air. Calculate the force of attraction between the charges.

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Asked on July 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-11T21:03:50.819Z

Here are the solutions to the questions. Question 6: 6(a)(i): State the laws of electrostatics. The two fundamental laws of electrostatics are: • Coulomb's Law: States that the force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The force acts along the line joining the charges. • Law of Conservation of Charge: States that the net electric charge of an isolated system remains constant. Charge can neither be created nor destroyed, only transferred. 6(a)(ii): Two electric charges of +15\, and +10\, are spaced 30\,cm apart in air. Calculate the force of attraction between the charges. Given: Charge q_1 = +15\, = 15 × 10^-6\,C Charge q_2 = +10\, = 10 × 10^-6\,C Distance r = 30\,cm = 0.3\,m Coulomb's constant k = 9 × 10^9\,N·m^2/C^2 Step 1: Apply Coulomb's Law formula. F = k (q_1 q_2)/(r^2) Step 2: Substitute the given values. F = (9 × 10^9\,N·m^2/C^2) (15 × 10^-6\,C)(10 × 10^-6\,C)(0.3\,m)^2 F = (9 × 10^9) 150 × 10^-120.09 F = (9 × 10^9) 1.5 × 10^-100.09 F = 13.5 × 10^-10.09 F = (1.35)/(0.09) F = 15\,N Since both charges are positive, the force is repulsive, not attractive. The force between the charges is 15 N (repulsive). 6(b): Calculate the phase currents and draw the vector diagram. Calculate the current in the neutral conductor. This question requires a circuit diagram or specific parameters (e.g., voltage, impedance, connection type) for a three-phase AC circuit, which are not provided in the image. Therefore, this question cannot be solved. 6(c): For the diagram shown above, calculate the charge and voltage associated with each of the changes in the circuit. Show your answers in a tabular form. The circuit shows a 30V DC source connected to a network of capacitors. The capacitors are: C_1 = 11\,, C_2 = 12\,, C_3 = 13\,, C_4 = 14\,, C_5 = 15\,, C_6 = 16\,. The circuit consists of three parallel branches, with each branch containing two capacitors in series. The total voltage across each parallel branch is V_total = 30\,V. Step 1: Calculate the equivalent capacitance for each series branch. For series capacitors, (1)/(C_eq) = (1)/(C_a) + (1)/(C_b). Branch 1 (C_1, C_2): (1)/(C_12) = (1)/(11\,) + (1)/(12\,) = (12+11)/(11 × 12)\,^-1 = (23)/(132)\,^-1 C_12 = (132)/(23)\, ≈ 5.739\, Branch 2 (C_3, C_4): (1)/(C_34) = (1)/(13\,) + (1)/(14\,) = (14+13)/(13 × 14)\,^-1 = (27)/(182)\,^-1 C_34 = (182)/(27)\, ≈ 6.741\, Branch 3 (C_5, C_6): (1)/(C_56) = (1)/(15\,) + (1)/(16\,) = (16+15)/(15 × 16)\,^-1 = (31)/(240)\,^-1 C_56 = (240)/(31)\, ≈ 7.742\, Step 2: Calculate the total charge for each series branch (Q = C_eq V_total). Charge in Branch 1 (Q_12): Q_12 = C_12 V_total = (132)/(23)\, × 30\,V = (3960)/(23)\, ≈ 172.17\, Charge in Branch 2 (Q_34): Q_34 = C_34 V_total = (182)/(27)\, × 30\,V = (5460)/(27)\, = (1820)/(9)\, ≈ 202.22\, Charge in Branch 3 (Q_56): Q_56 = C_56 V_total = (240)/(31)\, × 30\,V = (7200)/(31)\, ≈ 232.26\, In a series connection, the charge on each capacitor is the same as the total charge of the branch. Step 3: Calculate the voltage across each capacitor (V = Q/C). Branch 1: V_1 = Q_12C_1 = 3960/23\,11\, = (360)/(23)\,V ≈ 15.65\,V V_2 = Q_12C_2 = 3960/23\,12\, = (330)/(23)\,V ≈ 14.35\,V (Check: V_1 + V_2 = (360+330)/(23) = (690)/(23) = 30\,V) Branch 2: V_3 = Q_34C_3 = 1820/9\,13\, = (140)/(9)\,V ≈ 15.56\,V V_4 = Q_34C_4 = 1820/9\,14\, = (130)/(9)\,V ≈ 14.44\,V (Check: V_3 + V_4 = (140+130)/(9) = (270)/(9) = 30\,V) Branch 3: V_5 = Q_56C_5 = 7200/31\,15\, = (480)/(31)\,V ≈ 15.48\,V V_6 = Q_56C_6 = 7200/31\,16\, = (450)/(31)\,V ≈ 14.52\,V (Check: V_5 + V_6 = (480+450)/(31) = (930)/(31) = 30\,V) Step 4: Present the answers in a tabular form. | Capacitor | Capacitance (C) | Charge (Q) | Voltage (V) | | :-------- | :------------------ | :------------- | :------------ | | C_1 | 11\, | 172.17\, | 15.65\,V | | C_2 | 12\, | 172.17\, | 14.35\,V | | C_3 | 13\, | 202.22\, | 15.56\,V | | C_4 | 14\, | 202.22\, | 14.44\,V | | C_5 | 15\, | 232.26\, | 15.48\,V | | C_6 | 16\, | 232.26\, | 14.52\,V | Question 7: 7(a): Give a sketch of a two-winding transformer and label the main parts. A two-winding transformer consists of a primary winding and a secondary winding wound around a common laminated soft iron core. ` +---------------------------------+ | | Vp ---+ +-------------------------+ +--- Vs | | | | | | Laminated Iron Core | | | | | | Np ---+ +-------------------------+ +--- Ns | | +---------------------------------+ ` Primary Winding (Np): The coil connected to the input AC voltage source (V_p). Secondary Winding (Ns): The coil connected to the load, where the output voltage (V_s) is induced. Laminated Soft Iron Core: Provides a low reluctance path for the magnetic flux, linking the primary and secondary windings. It is laminated to reduce eddy current losses. 7(b): State the types of losses that occur in a transformer and explain how these losses can be minimized. The main types of losses in a transformer are: 1. Copper Losses (I²R Losses): These losses occur in the primary and secondary windings due to the resistance of the copper wire. They are proportional to the square of the current. Minimization*: Use thicker copper wires (lower resistance) for the windings and design the transformer for optimal current density. 2. Core Losses (Iron Losses): These losses occur in the transformer core and consist of two components: Hysteresis Loss*: Caused by the repeated magnetization and demagnetization of the core material as the AC current flows. Minimization*: Use core materials with low hysteresis loop area, such as silicon steel. Eddy Current Loss*: Caused by circulating currents induced in the core material due to the changing magnetic flux. Minimization*: Use a laminated core (thin sheets of core material insulated from each other) to increase the resistance to eddy current flow. 7(c): A two winding transformer has primary to secondary turn ratio 5:3. The input voltage is 400\,V, primary winding leakage resistance is 3\,, secondary winding resistance is 1\, and the load impedance is 6\,. (i) Draw circuit diagram of the transformer. The equivalent circuit diagram of a transformer referred to the primary side, assuming a purely resistive load: ` V1 ---[R1]-----[Ideal Transformer]-----[R2']-----[ZL']--- | | | | | | | | ------------------------ ` Where: V_1 = 400\,V (Input Voltage) R_1 = 3\, (Primary Winding Resistance) R_2' = Secondary Winding Resistance referred to primary Z_L' = Load Impedance referred to primary Ideal Transformer has turn ratio N_1:N_2 = 5:3. (ii) Calculate the Primary current, secondary current and secondary voltage. Given: Turn ratio (N_1)/(N_2) = (5)/(3) Input voltage V_1 = 400\,V Primary resistance R_1 = 3\, Secondary resistance R_2 = 1\, Load impedance Z_L = 6\, (assumed resistive) Step 1: Calculate the secondary resistance and load impedance referred to the primary side. R_2' = R_2 ((N_1)/(N_2))^2 = 1\, × ((5)/(3))^2 = 1 × (25)/(9) = (25)/(9)\, Z_L' = Z_L ((N_1)/(N_2))^2 = 6\, × ((5)/(3))^2 = 6 × (25)/(9) = (150)/(9) = (50)/(3)\, Step 2: Calculate the total equivalent impedance referred to the primary side. Z_eq,primary = R_1 + R_2' + Z_L' Z_eq,primary = 3 + (25)/(9) + (50)/(3) = (27)/(9) + (25)/(9) + (150)/(9) = (27+25+150)/(9) = (202)/(9)\, Step 3: Calculate the primary current (I_1). I_1 = (V_1)/(Z_eq,primary) = 400\,V202/9\, = (400 × 9)/(202) = (3600)/(202) = (1800)/(101)\,A I_1 ≈ 17.82\,A Step 4: Calculate the secondary current (I_2) using the current ratio. (I_1)/(I_2) = (N_2)/(N_1) I_2 = I_1 (N_1)/(N_2) I_2 = (1800)/(101)\,A × (5)/(3) = (600 × 5)/(101) = (3000)/(101)\,A I_2 ≈ 29.70\,A Step 5: Calculate the secondary voltage (V_2). V_2 = I_2 Z_L = (3000)/(101)\,A × 6\, = (18000)/(101)\,V V_2 ≈ 178.22\,V Primary current I_1 = (1800)/(101)\,A ≈ 17.82\,A Secondary current I_2 = (3000)/(101)\,A ≈ 29.70\,A Secondary voltage V_2 = (18000)/(101)\,V ≈ 178.22\,V (iii) Calculate the input power, output power and the efficiency of the transformer. Step 1: Calculate the input power (P_in). Since the load is resistive, the power factor is 1. P_in = V_1 I_1 = 400\,V × (1800)/(101)\,A = (720000)/(101)\,W P_in ≈ 7128.71\,W Step 2: Calculate the output power (P_out). P_out = V_2 I_2 = (18000)/(101)\,V × (3000)/(101)\,A = (54000000)/(10201)\,W P_out ≈ 5293.60\,W Alternatively, P_out = I_2^2 Z_L = ((3000)/(101))^2 × 6 = (9000000)/(10201) × 6 = (54000000)/(10201)\,W. Step 3: Calculate the efficiency (). = P_outP_in × 100\% = 54000000/10201\,W720000/101\,W × 100\% = (54000000)/(10201) × (101)/(720000) × 100\% = (54000000 × 101)/(101 × 101 × 720000) × 100\% = (54000000)/(101 × 720000) × 100\% = (540)/(101 × 7.2) × 100\% = (540)/(727.2) × 100\% ≈ 74.25\% Input power ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._

Accepted Answer · 2026-07-11T21:03:50.819Z

Here are the solutions to the questions. Question 6: 6(a)(i): State the laws of electrostatics. The two fundamental laws of electrostatics are: • Coulomb's Law: States that the force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The force acts along the line joining the charges. • Law of Conservation of Charge: States that the net electric charge of an isolated system remains constant. Charge can neither be created nor destroyed, only transferred. 6(a)(ii): Two electric charges of +15\, and +10\, are spaced 30\,cm apart in air. Calculate the force of attraction between the charges. Given: Charge q_1 = +15\, = 15 × 10^-6\,C Charge q_2 = +10\, = 10 × 10^-6\,C Distance r = 30\,cm = 0.3\,m Coulomb's constant k = 9 × 10^9\,N·m^2/C^2 Step 1: Apply Coulomb's Law formula. F = k (q_1 q_2)/(r^2) Step 2: Substitute the given values. F = (9 × 10^9\,N·m^2/C^2) (15 × 10^-6\,C)(10 × 10^-6\,C)(0.3\,m)^2 F = (9 × 10^9) 150 × 10^-120.09 F = (9 × 10^9) 1.5 × 10^-100.09 F = 13.5 × 10^-10.09 F = (1.35)/(0.09) F = 15\,N Since both charges are positive, the force is repulsive, not attractive. The force between the charges is 15 N (repulsive). 6(b): Calculate the phase currents and draw the vector diagram. Calculate the current in the neutral conductor. This question requires a circuit diagram or specific parameters (e.g., voltage, impedance, connection type) for a three-phase AC circuit, which are not provided in the image. Therefore, this question cannot be solved. 6(c): For the diagram shown above, calculate the charge and voltage associated with each of the changes in the circuit. Show your answers in a tabular form. The circuit shows a 30V DC source connected to a network of capacitors. The capacitors are: C_1 = 11\,, C_2 = 12\,, C_3 = 13\,, C_4 = 14\,, C_5 = 15\,, C_6 = 16\,. The circuit consists of three parallel branches, with each branch containing two capacitors in series. The total voltage across each parallel branch is V_total = 30\,V. Step 1: Calculate the equivalent capacitance for each series branch. For series capacitors, (1)/(C_eq) = (1)/(C_a) + (1)/(C_b). Branch 1 (C_1, C_2): (1)/(C_12) = (1)/(11\,) + (1)/(12\,) = (12+11)/(11 × 12)\,^-1 = (23)/(132)\,^-1 C_12 = (132)/(23)\, ≈ 5.739\, Branch 2 (C_3, C_4): (1)/(C_34) = (1)/(13\,) + (1)/(14\,) = (14+13)/(13 × 14)\,^-1 = (27)/(182)\,^-1 C_34 = (182)/(27)\, ≈ 6.741\, Branch 3 (C_5, C_6): (1)/(C_56) = (1)/(15\,) + (1)/(16\,) = (16+15)/(15 × 16)\,^-1 = (31)/(240)\,^-1 C_56 = (240)/(31)\, ≈ 7.742\, Step 2: Calculate the total charge for each series branch (Q = C_eq V_total). Charge in Branch 1 (Q_12): Q_12 = C_12 V_total = (132)/(23)\, × 30\,V = (3960)/(23)\, ≈ 172.17\, Charge in Branch 2 (Q_34): Q_34 = C_34 V_total = (182)/(27)\, × 30\,V = (5460)/(27)\, = (1820)/(9)\, ≈ 202.22\, Charge in Branch 3 (Q_56): Q_56 = C_56 V_total = (240)/(31)\, × 30\,V = (7200)/(31)\, ≈ 232.26\, In a series connection, the charge on each capacitor is the same as the total charge of the branch. Step 3: Calculate the voltage across each capacitor (V = Q/C). Branch 1: V_1 = Q_12C_1 = 3960/23\,11\, = (360)/(23)\,V ≈ 15.65\,V V_2 = Q_12C_2 = 3960/23\,12\, = (330)/(23)\,V ≈ 14.35\,V (Check: V_1 + V_2 = (360+330)/(23) = (690)/(23) = 30\,V) Branch 2: V_3 = Q_34C_3 = 1820/9\,13\, = (140)/(9)\,V ≈ 15.56\,V V_4 = Q_34C_4 = 1820/9\,14\, = (130)/(9)\,V ≈ 14.44\,V (Check: V_3 + V_4 = (140+130)/(9) = (270)/(9) = 30\,V) Branch 3: V_5 = Q_56C_5 = 7200/31\,15\, = (480)/(31)\,V ≈ 15.48\,V V_6 = Q_56C_6 = 7200/31\,16\, = (450)/(31)\,V ≈ 14.52\,V (Check: V_5 + V_6 = (480+450)/(31) = (930)/(31) = 30\,V) Step 4: Present the answers in a tabular form. | Capacitor | Capacitance (C) | Charge (Q) | Voltage (V) | | :-------- | :------------------ | :------------- | :------------ | | C_1 | 11\, | 172.17\, | 15.65\,V | | C_2 | 12\, | 172.17\, | 14.35\,V | | C_3 | 13\, | 202.22\, | 15.56\,V | | C_4 | 14\, | 202.22\, | 14.44\,V | | C_5 | 15\, | 232.26\, | 15.48\,V | | C_6 | 16\, | 232.26\, | 14.52\,V | Question 7: 7(a): Give a sketch of a two-winding transformer and label the main parts. A two-winding transformer consists of a primary winding and a secondary winding wound around a common laminated soft iron core. ` +---------------------------------+ | | Vp ---+ +-------------------------+ +--- Vs | | | | | | Laminated Iron Core | | | | | | Np ---+ +-------------------------+ +--- Ns | | +---------------------------------+ ` Primary Winding (Np): The coil connected to the input AC voltage source (V_p). Secondary Winding (Ns): The coil connected to the load, where the output voltage (V_s) is induced. Laminated Soft Iron Core: Provides a low reluctance path for the magnetic flux, linking the primary and secondary windings. It is laminated to reduce eddy current losses. 7(b): State the types of losses that occur in a transformer and explain how these losses can be minimized. The main types of losses in a transformer are: 1. Copper Losses (I²R Losses): These losses occur in the primary and secondary windings due to the resistance of the copper wire. They are proportional to the square of the current. Minimization*: Use thicker copper wires (lower resistance) for the windings and design the transformer for optimal current density. 2. Core Losses (Iron Losses): These losses occur in the transformer core and consist of two components: Hysteresis Loss*: Caused by the repeated magnetization and demagnetization of the core material as the AC current flows. Minimization*: Use core materials with low hysteresis loop area, such as silicon steel. Eddy Current Loss*: Caused by circulating currents induced in the core material due to the changing magnetic flux. Minimization*: Use a laminated core (thin sheets of core material insulated from each other) to increase the resistance to eddy current flow. 7(c): A two winding transformer has primary to secondary turn ratio 5:3. The input voltage is 400\,V, primary winding leakage resistance is 3\,, secondary winding resistance is 1\, and the load impedance is 6\,. (i) Draw circuit diagram of the transformer. The equivalent circuit diagram of a transformer referred to the primary side, assuming a purely resistive load: ` V1 ---[R1]-----[Ideal Transformer]-----[R2']-----[ZL']--- | | | | | | | | ------------------------ ` Where: V_1 = 400\,V (Input Voltage) R_1 = 3\, (Primary Winding Resistance) R_2' = Secondary Winding Resistance referred to primary Z_L' = Load Impedance referred to primary Ideal Transformer has turn ratio N_1:N_2 = 5:3. (ii) Calculate the Primary current, secondary current and secondary voltage. Given: Turn ratio (N_1)/(N_2) = (5)/(3) Input voltage V_1 = 400\,V Primary resistance R_1 = 3\, Secondary resistance R_2 = 1\, Load impedance Z_L = 6\, (assumed resistive) Step 1: Calculate the secondary resistance and load impedance referred to the primary side. R_2' = R_2 ((N_1)/(N_2))^2 = 1\, × ((5)/(3))^2 = 1 × (25)/(9) = (25)/(9)\, Z_L' = Z_L ((N_1)/(N_2))^2 = 6\, × ((5)/(3))^2 = 6 × (25)/(9) = (150)/(9) = (50)/(3)\, Step 2: Calculate the total equivalent impedance referred to the primary side. Z_eq,primary = R_1 + R_2' + Z_L' Z_eq,primary = 3 + (25)/(9) + (50)/(3) = (27)/(9) + (25)/(9) + (150)/(9) = (27+25+150)/(9) = (202)/(9)\, Step 3: Calculate the primary current (I_1). I_1 = (V_1)/(Z_eq,primary) = 400\,V202/9\, = (400 × 9)/(202) = (3600)/(202) = (1800)/(101)\,A I_1 ≈ 17.82\,A Step 4: Calculate the secondary current (I_2) using the current ratio. (I_1)/(I_2) = (N_2)/(N_1) I_2 = I_1 (N_1)/(N_2) I_2 = (1800)/(101)\,A × (5)/(3) = (600 × 5)/(101) = (3000)/(101)\,A I_2 ≈ 29.70\,A Step 5: Calculate the secondary voltage (V_2). V_2 = I_2 Z_L = (3000)/(101)\,A × 6\, = (18000)/(101)\,V V_2 ≈ 178.22\,V Primary current I_1 = (1800)/(101)\,A ≈ 17.82\,A Secondary current I_2 = (3000)/(101)\,A ≈ 29.70\,A Secondary voltage V_2 = (18000)/(101)\,V ≈ 178.22\,V (iii) Calculate the input power, output power and the efficiency of the transformer. Step 1: Calculate the input power (P_in). Since the load is resistive, the power factor is 1. P_in = V_1 I_1 = 400\,V × (1800)/(101)\,A = (720000)/(101)\,W P_in ≈ 7128.71\,W Step 2: Calculate the output power (P_out). P_out = V_2 I_2 = (18000)/(101)\,V × (3000)/(101)\,A = (54000000)/(10201)\,W P_out ≈ 5293.60\,W Alternatively, P_out = I_2^2 Z_L = ((3000)/(101))^2 × 6 = (9000000)/(10201) × 6 = (54000000)/(10201)\,W. Step 3: Calculate the efficiency (). = P_outP_in × 100\% = 54000000/10201\,W720000/101\,W × 100\% = (54000000)/(10201) × (101)/(720000) × 100\% = (54000000 × 101)/(101 × 101 × 720000) × 100\% = (54000000)/(101 × 720000) × 100\% = (540)/(101 × 7.2) × 100\% = (540)/(727.2) × 100\% ≈ 74.25\% Input power ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._

State the laws of electrostatics. Two electric charges of +15µC and +1µC are spaced 50 cm apart in air. Calculate the force of attraction between the charges.

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State the laws of electrostatics. Two electric charges of +15µC and +1µC are spaced 50 cm apart in air. Calculate the force of attraction between the charges.

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