Question

The electric field produced by a positively charged particle, placed in x-y plane is $1.8(3\hat{i}+4\hat{j})$ N/C at (5m ; 5m) and $1.8(4\hat{i}+3\hat{j})$ N/C at (6m, 4m). Then :

• A. Charged particle is at origin.
• B. Charged particle is at (2m, 1m).
• C. Magnitude of charge is $2.5\times {10}^{-8}C$.
• D. The electric potential at (-1m, -3m) is 45V.

Answer 1

Answer: (B), (C), (D)

The correct options are
B Charged particle is at (2m, 1m).
C Magnitude of charge is $2.5\times {10}^{-8}C$.
D The electric potential at (-1m, -3m) is 45V.

$\frac{x-5}{y-5}=\frac{3}{4}\Rightarrow 4x-3y=5$ $\cdots$ (i)
$\frac{x-6}{y-4}=\frac{4}{3}\Rightarrow 3x-4y=2$ $\cdots$ (ii)
Solve (i) and (ii)
x = 2; y = 1
Position vector of (5m, 5m) w.r.t. charged particle $=(5-2)\hat{i}+(5-1)\hat{j}=3\hat{i}+4\hat{j}$
Now, $E=\frac{9\times {10}^9\times Q}{|3\hat{i}+4\hat{j}{|}^2}\Rightarrow Q=2.5\times {10}^{-3}C$; Similarly $V=\frac{KQ}{r}=45V$