Question

At any point $ \left( x,0,0 \right) $ the electric potential $ V $ is $ \left( \dfrac{1000}{x}+\dfrac{1500}{{{x}^{2}}}+\dfrac{500}{{{x}^{3}}} \right)\text{Volt} $ then electric field intensity at $ x=1m $ is
(a) $ 5500\left( \hat{j}+\hat{k} \right)\text{V/m} $
(b) $ 5500\text{ iV/m} $
(c) $ \dfrac{5500}{\sqrt{2}}\left( \hat{j}+\hat{k} \right)\text{V/m} $
(d) $ \dfrac{5500}{\sqrt{2}}\left( \hat{i}+\hat{k} \right)\text{V/m} $

Answer 1

Hint :Electric field intensity depends on rate of change electric potential with distance. So, we need to calculate rate of change of electric potential which is $ \dfrac{dv}{dr} $
Formula used:
 $ E=\dfrac{-dv}{dr} $
Where,
 $ E= $ Electric field intensity
 $ -\dfrac{-dv}{dr}= $ rate of change of electric potential with distance.

Complete Step By Step Answer:
As per data given in the question we have,
 $ V=\left( \dfrac{1000}{x}+\dfrac{1500}{{{x}^{2}}}+\dfrac{500}{{{x}^{3}}} \right)\text{Volt} $
At $ x $ direction distance $ \left( x \right)=1m $
As we know that $ E=\dfrac{-dv}{dr} $ , So, electric field intensity in $ x $ direction becomes $ E=\dfrac{-dv}{dx}\hat{i}...(i) $
So, now we find derivative of $ V $
 $ \dfrac{dv}{dx}=\dfrac{d}{dx}=\left( \dfrac{1000}{x}+\dfrac{1500}{{{x}^{2}}}+\dfrac{500}{{{x}^{3}}} \right) $
 $ =1000\left( \dfrac{-1}{{{x}^{2}}} \right)+1500\left( \dfrac{-2}{{{x}^{3}}} \right)+500\left( \dfrac{-3}{{{x}^{4}}} \right) $
Here $ x=1m, $ then
 $ \dfrac{dv}{dx}=-\left( 1000+3000+1500 \right) $
 $ \dfrac{dv}{dx}=-55000 $
Now, we will put this value of $ \dfrac{dv}{dx} $ in electric field intensity formula,
 $ {{E}_{x=1}}=\dfrac{-dv}{dx}\hat{i} $
 $ =-\left( -5500 \right)\hat{i} $
 $ {{E}_{x=1}}=5500\hat{i}\text{ V/m} $
Hence the option (b) is the correct answer.

Note :
Force experienced by a unit positive charge placed at a point is called electric field intensity. Electric field intensity is a vector quantity and it is denoted by $ E. $
Formula of electric field intensity is $ E=\dfrac{F}{Q} $
Unit of electric field intensity is $ N{{C}^{-1}}\text{ or V}{{\text{m}}^{-1}} $ due to positive charge electric field intensity is always directed away from the charge and due to Negative charge electric field intensity is always directed towards the charge.
Force experienced in an electric field is $ F=QE $ electric field intensity depends upon rate of charge of electric potential with distance.
 $ E=\dfrac{-dV}{dr} $
Pay attention to data given in question, Here electric potential is given with question not rate of change of electric potential.