Question

In a region electric field varies as $\overrightarrow{E}=(x\hat{i}-2y\hat{j}+z\hat{k})$. The potential difference between points $A[2m,1m,0m]$ and $B[0m,2m,4m]$ is:

• A. $V_{AB}=2V$
• B. $V_{AB}=3V$
• C. $V_{AB}=4V$
• D. $V_{AB}=9V$

Answer 1

The correct option is B $V_{AB}=3V$

Electric field varies as

$\overrightarrow{E}=(x\hat{i}-2h\hat{j}+z\hat{k})$

The potential difference

$A[2m,1m,0m]B[0m,2m,4m]$

The formula of the potential difference

$V=-E.d$

Where $E=$ Electric field

$d=$ distance between the electric field

$V=$ Potential difference

The distance $=\overrightarrow{AB}$

$=\sqrt{(2-0{)}^2+(1-2{)}^2+(0-4{)}^2}=\sqrt{4+1+16}=\sqrt{21}{V}_{AB}=\sqrt{21}(x\hat{i}+2y\hat{j}+z\hat{k})$

${\int }_B^{A}dV={\int }_{(0,2,4)}^{(2,1,0)}(x\hat{i}+2y\hat{j}+z\hat{k})(dx\hat{i}+dy\hat{j}+dz\hat{k})V_A-V_B=-{\int }_{(0,2,4)}^{(2,1,0)}(x\hat{i}+2y\hat{j}+z\hat{k})(dx\hat{i}+dy\hat{j}+dz\hat{k})=(2+1)V=3V$