Electrostatics Question 327
Question: The electric potential $ V $ at any point O (x, y, z all in metres) in space is given by $ V=4x^{2}volt $ . The electric field at the point $ (1m,0,2m) $ in $ volt/metre $ is [IIT 1992; RPET 1999; MP PMT 2001]
Options:
A) 8 along negative $ X- $ axis
B) 8 along positive $ X- $ axis
C) 16 along negative $ X- $ axis
D) 16 along positive $ Z- $ axis
Show Answer
Answer:
Correct Answer: A
Solution:
The electric potential $ V(x,y,z)=4x^{2}volt $
Now $ \overrightarrow{E}=-( \hat{i}\frac{\partial V}{\partial x}+\hat{j}\frac{\partial V}{\partial y}+\hat{k}\frac{\partial V}{\partial z} ) $ Now $ \frac{\partial V}{\partial x}=8x,\frac{\partial V}{\partial y}=0 $
and $ \frac{\partial V}{\partial z}=0 $
Hence $ \overrightarrow{E}=-8x\hat{i} $
, so at point (1m, 0, 2m) $ \overrightarrow{E}=-8\hat{i}volt/metre $ or 8 along negative X-axis.
