Two point charges 4μC and 9μCare separated by a distance of 50 cm.&#

Question

Two point charges

4 μ C

and

9 μ C

are separated by a distance of 50 cm. The potential at the point between them where the field has zero strength is

Options:

A . 4.5×105V

B . 9×105V

C . 9×104V

D . Zero

Answer: Option A
:
A
Assume the point at which field is zero is at a distance X from

4 μ C

X = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} + 1} = \frac{50}{\sqrt{\frac{9}{4}} + 1} = \frac{50}{\frac{3}{2} + 1} = \frac{100}{5} = 20 c m V = V_{1} + V_{2} = \frac{1}{4 π ϵ_{0}} [\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}] = 9 \times 10^{9} [\frac{4 \times 10^{- 6}}{0.2} + \frac{9 \times 10^{- 6}}{0.3}] = 9 \times 10^{3} [20 + 30] = 9 \times 10^{3} \times 50 = 4.5 \times 10^{5} V

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