A steady potential of 500V is maintained across two capacitors, C1 = 5.0mf and C2 = 2.0mf connected in series. Calculate i) the potential difference across each capacitor, ii) the charge on each capacitor, iii) the energy stored in each capacitor.

Question

Asked on July 20, 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 · 2026-07-20T07:45:53.675Z

Here's how to solve the problem: Given: Total potential difference V = 500 \, V Capacitance C_1 = 5.0 \, mF = 5.0 × 10^-3 \, F Capacitance C_2 = 2.0 \, mF = 2.0 × 10^-3 \, F The capacitors are connected in series. Step 1: Calculate the equivalent capacitance (C_eq) for capacitors in series. For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances. (1)/(C_eq) = (1)/(C_1) + (1)/(C_2) (1)/(C_eq) = (1)/(5.0 × 10^-3) \, F + (1)/(2.0 × 10^-3) \, F (1)/(C_eq) = 200 \, F^-1 + 500 \, F^-1 (1)/(C_eq) = 700 \, F^-1 C_eq = (1)/(700) \, F Step 2: Calculate the charge on each capacitor (ii). In a series connection, the charge (Q) on each capacitor is the same and equal to the total charge stored by the equivalent capacitance. Q = C_eq V Q = ((1)/(700) \, F) × (500 \, V) Q = (500)/(700) \, C = (5)/(7) \, C The charge on each capacitor is ii) (5)/(7) \, C ≈ 0.714 \, C. Step 3: Calculate the potential difference across each capacitor (i). The potential difference across each capacitor is given by V = (Q)/(C). For C_1: V_1 = (Q)/(C_1) V_1 = (5)/(7) \, C5.0 × 10^-3 \, F = (5)/(7 × 0.005) \, V = (5)/(0.035) \, V = (1000)/(7) \, V For C_2: V_2 = (Q)/(C_2) V_2 = (5)/(7) \, C2.0 × 10^-3 \, F = (5)/(7 × 0.002) \, V = (5)/(0.014) \, V = (2500)/(7) \, V The potential difference across each capacitor is: i) V_1 = (1000)/(7) \, V ≈ 142.86 \, V i) V_2 = (2500)/(7) \, V ≈ 357.14 \, V Step 4: Calculate the energy stored in each capacitor (iii). The energy stored in a capacitor is given by U = (1)/(2) Q V. For C_1: U_1 = (1)/(2) Q V_1 U_1 = (1)/(2) ((5)/(7) \, C) ((1000)/(7) \, V) U_1 = (5000)/(98) \, J = (2500)/(49) \, J For C_2: U_2 = (1)/(2) Q V_2 U_2 = (1)/(2) ((5)/(7) \, C) ((2500)/(7) \, V) U_2 = (12500)/(98) \, J = (6250)/(49) \, J The energy stored in each capacitor is: iii) U_1 = (2500)/(49) \, J ≈ 51.02 \, J iii) U_2 = (6250)/(49) \, J ≈ 127.55 \, J Send me the next one 📸

Accepted Answer · 2026-07-20T07:45:53.675Z

Here's how to solve the problem: Given: Total potential difference V = 500 \, V Capacitance C_1 = 5.0 \, mF = 5.0 × 10^-3 \, F Capacitance C_2 = 2.0 \, mF = 2.0 × 10^-3 \, F The capacitors are connected in series. Step 1: Calculate the equivalent capacitance (C_eq) for capacitors in series. For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances. (1)/(C_eq) = (1)/(C_1) + (1)/(C_2) (1)/(C_eq) = (1)/(5.0 × 10^-3) \, F + (1)/(2.0 × 10^-3) \, F (1)/(C_eq) = 200 \, F^-1 + 500 \, F^-1 (1)/(C_eq) = 700 \, F^-1 C_eq = (1)/(700) \, F Step 2: Calculate the charge on each capacitor (ii). In a series connection, the charge (Q) on each capacitor is the same and equal to the total charge stored by the equivalent capacitance. Q = C_eq V Q = ((1)/(700) \, F) × (500 \, V) Q = (500)/(700) \, C = (5)/(7) \, C The charge on each capacitor is ii) (5)/(7) \, C ≈ 0.714 \, C. Step 3: Calculate the potential difference across each capacitor (i). The potential difference across each capacitor is given by V = (Q)/(C). For C_1: V_1 = (Q)/(C_1) V_1 = (5)/(7) \, C5.0 × 10^-3 \, F = (5)/(7 × 0.005) \, V = (5)/(0.035) \, V = (1000)/(7) \, V For C_2: V_2 = (Q)/(C_2) V_2 = (5)/(7) \, C2.0 × 10^-3 \, F = (5)/(7 × 0.002) \, V = (5)/(0.014) \, V = (2500)/(7) \, V The potential difference across each capacitor is: i) V_1 = (1000)/(7) \, V ≈ 142.86 \, V i) V_2 = (2500)/(7) \, V ≈ 357.14 \, V Step 4: Calculate the energy stored in each capacitor (iii). The energy stored in a capacitor is given by U = (1)/(2) Q V. For C_1: U_1 = (1)/(2) Q V_1 U_1 = (1)/(2) ((5)/(7) \, C) ((1000)/(7) \, V) U_1 = (5000)/(98) \, J = (2500)/(49) \, J For C_2: U_2 = (1)/(2) Q V_2 U_2 = (1)/(2) ((5)/(7) \, C) ((2500)/(7) \, V) U_2 = (12500)/(98) \, J = (6250)/(49) \, J The energy stored in each capacitor is: iii) U_1 = (2500)/(49) \, J ≈ 51.02 \, J iii) U_2 = (6250)/(49) \, J ≈ 127.55 \, J Send me the next one 📸

A steady potential of 500V is maintained across two capacitors, C1 = 5.0mf and C2 = 2.0mf connected in series. Calculate i) the potential difference across each capacitor, ii) the charge on each capacitor, iii) the energy stored in each capacitor.

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A steady potential of 500V is maintained across two capacitors, C1 = 5.0mf and C2 = 2.0mf connected in series. Calculate i) the potential difference across each capacitor, ii) the charge on each capacitor, iii) the energy stored in each capacitor.

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