Electrostatics Question 774
Question: The potential at the point x (measured in$ \mu m $ ) due to some charges situated on the x-axis is given by $ V(x)=20(x^{2}-4) $ volt. The electric field E at $ x=4\mu m $
Options:
A) $ ( 10/9 )volt/\mu m $ and in the +ve x direction
B) $ ( 5/3 )volt/\mu m $ and in the -ve x direction
C) $ ( 5/3 )volt/\mu m $ and in the +ve x direction
D) $ ( 10/9 )volt/\mu m $ and in the -ve x direction
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Here, $ V( x )=\frac{20}{x^{2}-4}volt $
We know that $ E=-\frac{dV}{dx}=-\frac{d}{dx}( \frac{20}{x^{2}-4} ) $
$ \text{or, }E=+\frac{40x}{{{( x^{2}-4 )}^{2}}} $
At $ x=4\mu m $ ,
$ E=+\frac{40\times 4}{{{(4^{2}-4)}^{2}}}=+\frac{160}{144}=+\frac{10}{9}volt/\mu m $ .
Positive sign indicates that $ \vec{E} $ is in +ve x-direction.
