Which statement is correct regarding the force between two charges q1 and q2 separated by a distance d. According to Coulomb's Law, the electrostatic force (F) between two point charges is d

Step 1: Solve Question 319.
The question asks which statement is correct regarding the force between two charges $q_1$ and $q_2$ separated by a distance $d$. According to Coulomb's Law, the electrostatic force ($F$) 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 correct option is (গ) $F \propto \frac{q_1 q_2}{d^2}$.

Step 2: Solve Question 320.
The question asks which of the given statements is correct regarding fundamental forces.
Statement i: মহাকর্ষ বল একটি দুর্বল বল। (Gravitational force is a weak force.) This is true; it is the weakest fundamental force.
Statement iv: সবল নিউক্লিয় বলের পাল্লা খুব কম। (The range of strong nuclear force is very short.) This is also true; the strong nuclear force acts over very short distances within the nucleus.

• The correct option is (ঘ) i ও iv.

Step 3: Solve Question 321.
The question asks for the unit of Coulomb's constant ($k$). From Coulomb's Law, $F = k \frac{q_1 q_2}{d^2}$. Rearranging for $k$, we get $k = \frac{F d^2}{q_1 q_2}$.
Substituting the units:

$$k = \frac{\text{N} \cdot \text{m}^2}{\text{C} \cdot \text{C}} = \text{N} \cdot \text{m}^2 \cdot \text{C}^{-2}$$

• The correct option is (খ) $\boxed{\text{Nm}^2\text{C}^{-2}}$.

Step 4: Solve Question 322.
The question asks for the electric field strength ($E$) when a 5 C charge experiences a 1.5 N force. Electric field strength is defined as force per unit charge.

$$E = \frac{F}{q}$$

Substitute the given values:

$$E = \frac{1.5 \text{ N}}{5 \text{ C}} = 0.3 \text{ N/C}$$

Among the given options, 0.33 N/C is the closest value.

• The correct option is (ঘ) $\boxed{0.33 \text{ N/C}}$.

Step 5: Solve Question 323.
First, calculate the electric field strength ($E$) at the point where a 40 C charge experiences a 160 N force.

$$E = \frac{F_1}{q_1} = \frac{160 \text{ N}}{40 \text{ C}} = 4 \text{ N/C}$$

Now, calculate the force ($F_2$) on a 50 C charge ($q_2$) at the same point.

$$F_2 = E \times q_2 = 4 \text{ N/C} \times 50 \text{ C} = 200 \text{ N}$$

• The correct option is (ঘ) $\boxed{200}$.

Step 6: Solve Question 324.
The question asks for the electric potential ($V$) at a point if 800 J of work ($W$) is done to move a 16 C charge ($q$) from a zero potential reference point to that point. Electric potential is defined as work done per unit charge.

$$V = \frac{W}{q}$$

Substitute the given values:

$$V = \frac{800 \text{ J}}{16 \text{ C}} = 50 \text{ V}$$

• The correct option is (খ) $\boxed{50 \text{ V}}$.

Step 7: Solve Question 325.
Given a capacitor with capacitance $C = 25 \text{ } \mu\text{F} = 25 \times 10^{-6} \text{ F}$ and charge $Q = 3.5 \times 10^{-4} \text{ C}$.
i. ক্যাপাসিটরটি চার্জিত আছে। (The capacitor is charged.) This is true because there is a charge $Q$ on its plates.
ii. এর বিভব 14 V। (Its potential is 14 V.) Calculate potential $V = \frac{Q}{C}$.

$$V = \frac{3.5 \times 10^{-4} \text{ C}}{25 \times 10^{-6} \text{ F}} = \frac{3.5}{25} \times 10^2 \text{ V} = 0.14 \times 100 \text{ V} = 14 \text{ V}$$

This statement is true.
iii. এর সঞ্চিত শক্তি 2.45 mJ। (Its stored energy is 2.45 mJ.) Calculate stored energy $U = \frac{1}{2}QV$.

$$$$