MP Board · Class 12 · Physics · Electric Charges and FieldsTwo point charges $q1 = +5 \times 10^{-9} \text{ C}$ and $q2 = -3 \times 10^{-9} \text{ C}$ are located $0.4 \text{ m}$ apart in vacuum. (a) What is the electric field at the midpoint $O$ of the line joining the two charges? (b) If a negative test charge of magnitude $q3 = -1.5 \times 10^{-9} \text{ C}$ is placed at this midpoint, what is the force experienced by the test charge?

Given Data:

• Charge $q_1 = +5 \times 10^{-9} \text{ C}$
• Charge $q_2 = -3 \times 10^{-9} \text{ C}$
• Total distance between charges, $d = 0.4 \text{ m}$
• Distance of midpoint $O$ from each charge, $r = \frac{d}{2} = \frac{0.4}{2} = 0.2 \text{ m}$
• Coulomb's constant, $k = \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$

Part (a): Electric Field at the Midpoint $O$

1. Magnitude of Electric Field due to $q_1$ ($E_1$): $$E_1 = \frac{k \cdot |q_1|}{r^2}$$ $$E_1 = \frac{(9 \times 10^9) \times (5 \times 10^{-9})}{(0.2)^2}$$ $$E_1 = \frac{45}{0.04} = 1125 \text{ N/C}$$ Direction: Since $q_1$ is positive, the electric field $\vec{E}_1$ at point $O$ points away from $q_1$ (towards $q_2$).

2. Magnitude of Electric Field due to $q_2$ ($E_2$): $$E_2 = \frac{k \cdot |q_2|}{r^2}$$ $$E_2 = \frac{(9 \times 10^9) \times (3 \times 10^{-9})}{(0.2)^2}$$ $$E_2 = \frac{27}{0.04} = 675 \text{ N/C}$$ Direction: Since $q_2$ is negative, the electric field $\vec{E}_2$ at point $O$ points towards $q_2$ (in the same direction as $\vec{E}_1$).

3. Net Electric Field ($E_{\text{net}}$): Since both electric fields $\vec{E}1$ and $\vec{E}2$ act in the same direction (from $q_1$ towards $q_2$), they add up algebraically: $$E{\text{net}} = E_1 + E_2$$ $$E{\text{net}} = 1125 + 675 = 1800 \text{ N/C}$$ Direction of $E_{\text{net}}$: Directed from $q_1$ to $q_2$.

Part (b): Force Experienced by the Test Charge $q_3$

1. Given Test Charge: $$q_3 = -1.5 \times 10^{-9} \text{ C}$|

2. Formula for Force: $$F = q_3 \cdot E_{\text{net}}$|

3. Calculation: $$F = (1.5 \times 10^{-9} \text{ C}) \times (1800 \text{ N/C})$$ $$F = 2700 \times 10^{-9} \text{ N} = 2.7 \times 10^{-6} \text{ N}$|

4. Direction of Force: Since $q_3$ is a negative charge, the force acts in a direction opposite to the net electric field. Therefore, the force is directed from $q_2$ towards $q_1$.

Final Answer:

• (a) The net electric field at the midpoint is $1800 \text{ N/C}$ directed from $q_1$ to $q_2$.
• (b) The force experienced by the test charge is $2.7 \times 10^{-6} \text{ N}$ directed from $q_2$ to $q_1$.