Question

Two tiny spheres carrying charges 1.8 $\mu C$ and $2.8\mu C$ are located at 40 cm apart. The potential at a point 20 cm from the mid – point of the line joining the two charges in a plane normal to the line and passing through the mid – point is.

• A. $1.4 \times 10^{5}V$
• B. $4.2 \times 10^{3}V$
• C. $2.9 \times 10^{4}V$
• D. $3.7 \times 10^{5}V$

Answer 1

Answer: (A)

$1.4 \times 10^{5}V$

Answer 2

: From the figure, potential at P,

$V = \frac{q_{1}}{4\pi\varepsilon_{0}(PA)} + \frac{q_{2}}{4\pi\varepsilon_{0}(PB)}$

$= \frac{1}{4\pi\varepsilon_{0}(PA)}(q_{1} + q_{2})$

$\because PA = PB = \sqrt{(0.2)^{2} + (0.2)^{2}} = 0.28m$)

$= \frac{9 \times 10^{9}(1.8 + 2.8) \times 10^{- 6}}{0.28} = 1.4 \times 10^{5}V$