The potential difference between two points $A(2,1,0) \mathrm{m}$ and $B(0,2,4) \mathrm{m}$ in an electric…
The potential difference between two points $A(2,1,0) \mathrm{m}$ and $B(0,2,4) \mathrm{m}$ in an electric field $(x \hat{\mathbf{i}}-2 y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \mathrm{Vm}^{-1}$ is
2 V
3 V
1 V
6 V
Solution
Given, electric field $\mathbf{E}=(x \hat{\mathbf{i}}-2 y \hat{\mathbf{i}}+z \hat{\mathbf{k}}) \mathrm{Vm}^{-1}$
Now, electric potential difference between two points $A(2,1,0)$ and $B(0,2,4)$ is given as
$
\begin{aligned}
& \therefore \quad \Delta V=-\int \mathbf{E} \cdot \mathbf{d r} \\
& \text { or }-\int(x \hat{\mathbf{i}}-2 y \hat{\mathbf{j}}+z \hat{\mathbf{k}})(d x \hat{\mathbf{i}}+d y \hat{\mathbf{j}}+d z \hat{\mathbf{k}}) \\
& =-\int_2^0 x d x+2 \int_1^2 y d y-\int_0^4 z d z \\
& \Rightarrow \quad\left[\frac{-x^2}{2}\right]_2^0+\left[y^2\right]^2-\left[\frac{z^2}{2}\right]_0^4 \\
& \Rightarrow \quad \frac{-1}{2}\left(0^2-2^2\right)+\left(2^2-1^2\right)-\frac{1}{2}\left(4^2-0^2\right) \\
&
\end{aligned}
$
$\begin{gathered}\Rightarrow \quad-\frac{1}{2}(-4)+(4-1)-\frac{1}{2}(16)=+2+3-8 \\ \Delta V=-3 \mathrm{~V} \text { or }|\Delta V|=3 \mathrm{~V}\end{gathered}$