The electric potential and electric field at a point due to a point charge are 1800 V and 900 N / C respectively. Then magnitude of the point charge should be :-

Question

The electric potential and electric field at a point due to a point charge are 1800 V and 900 N / C respectively. Then magnitude of the point charge should be :- (A) 3 μ C (B) 0.4 μ C (C) 0.2 μ C (D) 0.5 μ C

Tardigrade Admin · Accepted Answer

V =( KQ / r ) E =( KQ / r 2) ( E / V )=(1/ r ) ⇒ (900/1800)=(1/ r ) r =2 m 1800=(9 × 109 × Q /2) ⇒ Q =0.4 μ C

Q. The electric potential and electric field at a point due to a point charge are $1800 V$ and $900 N / C$ respectively. Then magnitude of the point charge should be :-

$3 μ C$

$0.4 μ C$

$0.2 μ C$

$0.5 μ C$

$V = \frac{K Q}{r}$
$E = \frac{K Q}{r ^{2}}$
$\frac{E}{V} = \frac{1}{r}$
$\Rightarrow \frac{900}{1800} = \frac{1}{r}$
$r = 2 m$
$1800 = \frac{9 \times 1 0 ^{9} \times Q}{2}$
$\Rightarrow Q = 0.4 μ C$

Q. The electric potential and electric field at a point due to a point charge are 1800V and 900N/C respectively. Then magnitude of the point charge should be :-

3μC

0.4μC

0.2μC

0.5μC