Electrostatic Potential and Capacitance - Result Question 4
4. If potential (in volts) in a region is expressed as $V(x, y, z)=6 x y-y+2 y z$, the electric field (in N/C) at point $(1,1,0)$ is :
[2015 RS]
(a) $-(6 \hat{i}+5 \hat{j}+2 \hat{k})$
(b) $-(2 \hat{i}+3 \hat{j}+\hat{k})$
(c) $-(6 \hat{i}+9 \hat{j}+\hat{k})$
(d) $-(3 \hat{i}+5 \hat{j}+3 \hat{k})$
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Answer:
Correct Answer: 4. (a)
Solution:
- (a) Potential in a region is given by,
$V=6 xy-y+2 yz$
As we know the relation between electric potential and electric field is
$\vec{E}=-(\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \hat{k})$
where $V=6 x y-y+2 y z$
$\vec{E}=-[(6 y \hat{i}+(6 x-1+2 z) \hat{j}+(2 y) \hat{k}]$
$\overrightarrow{{}E} _{(1,1,0)}=-(6 \hat{i}+5 \hat{j}+2 \hat{k})$
With the help of formula $E=\frac{-d v}{d r}$, potential difference between any two point in an electric field can be determined by knowing the boundary
$ \text{ conditions } d V=-\int _{\eta_1}^{r_2} \vec{E} \overrightarrow{{}d r}=-\int _{\eta_1}^{r_2} E \cdot d r \cos \theta $
(d) $\overrightarrow{{}E}=-\frac{\partial V}{\partial x} \hat{i}-\frac{\partial V}{\partial y} \hat{j}-\frac{\partial V}{\partial z} \hat{k}$
where $V(x, y, z)=6 x-8 x y-8 y+6 y z$
$=-[(6-8 y) \hat{i}+(-8 x-8+6 z) \hat{j}+(6 y) \hat{k}]$
At $(1,1,1), \vec{E}=2 \hat{i}+10 \hat{j}-6 \hat{k}$
$\Rightarrow \quad(\overrightarrow{{}E})=\sqrt{(2)^{2}+(10)^{2}+(-6)^{2}}=\sqrt{140}=2 \sqrt{35}$
$\therefore \quad F=q \overrightarrow{{}E}=2 \times 2 \sqrt{35}=4 \sqrt{35}$
