Question

Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}$ C and $-8 \times 10^{-6}$ C respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}$ NC$^{-1}$. The distance '$d$' between the point charges $A$ and $B$ is:

• A. 3.0 m
• B. 4.0 m
• C. 1.0 m
• D. 2.0 m

Answer 1

Answer: (A)

3.0 m

Answer 2

Let the two point charges be $q_A = +8 \times 10^{-6}$ C and $q_B = -8 \times 10^{-6}$ C. They are placed at a distance $d$ apart. The midpoint $O$ is located at a distance $r = d/2$ from both charges.

The electric field at point $O$ due to charge $q_A$ is directed away from $q_A$ (since $q_A$ is positive). At the midpoint $O$, this direction is towards charge $B$. The magnitude of this electric field is: $E_A = \frac{k|q_A|}{r^2} = \frac{k(8 \times 10^{-6})}{(d/2)^2}$

The electric field at point $O$ due to charge $q_B$ is directed towards $q_B$ (since $q_B$ is negative). At the midpoint $O$, this direction is also towards charge $B$. The magnitude of this electric field is: $E_B = \frac{k|q_B|}{r^2} = \frac{k|-8 \times 10^{-6}|}{(d/2)^2} = \frac{k(8 \times 10^{-6})}{(d/2)^2}$

Since both electric fields $E_A$ and $E_B$ are in the same direction (towards $B$), the net electric field at $O$ is the sum of their magnitudes: $E_{net} = E_A + E_B = \frac{k(8 \times 10^{-6})}{(d/2)^2} + \frac{k(8 \times 10^{-6})}{(d/2)^2} = 2 \times \frac{k(8 \times 10^{-6})}{(d/2)^2}$ $E_{net} = \frac{2k(8 \times 10^{-6})}{d^2/4} = \frac{8k(8 \times 10^{-6})}{d^2}$

We are given that the electric field at the midpoint $O$ is $6.4 \times 10^4$ NC$^{-1}$. We use the value of Coulomb's constant $k \approx 9 \times 10^9$ Nm$^2$C$^{-2}$. $6.4 \times 10^4 = \frac{8 \times (9 \times 10^9) \times (8 \times 10^{-6})}{d^2}$

Now, we solve for $d^2$: $d^2 = \frac{8 \times 9 \times 8 \times 10^{9-6}}{6.4 \times 10^4}$ $d^2 = \frac{576 \times 10^3}{6.4 \times 10^4}$ Rewrite $6.4 \times 10^4$ as $64 \times 10^3$: $d^2 = \frac{576 \times 10^3}{64 \times 10^3}$ $d^2 = \frac{576}{64}$

To simplify $\frac{576}{64}$: $\frac{576}{64} = \frac{9 \times 64}{64} = 9$

So, $d^2 = 9$. Taking the square root, we get $d = \sqrt{9} = 3$ m.

The distance '$d$' between the point charges $A$ and $B$ is 3.0 m.