Question 289: Electrostats & Capacitance

Question

Electric field in a region is given by $\vec{E}=A x \hat{i}+B y \hat{j}$, where $A=10 \mathrm{~V} / \mathrm{m}^2$ and $B=5 \mathrm{~V} / \mathrm{m}^2$. If the electric potential at a point $(10,20)$ is 500 V , then the electric potential at origin is $\_\_\_\_$ V.

Accepted Answer

Electric potential difference between two points is defined as the negative of the work done by the electric field in moving a unit positive charge from the initial point to the final point.
The infinitesimal work done $d W$ by an electric field $\vec{E}$ on a unit charge over a tiny displacement $d \vec{r}$ is $\vec{E} \cdot d \vec{r}$.
Therefore, the infinitesimal change in potential, dV, is :
$$ \mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $$
To find the total potential difference between an initial position $\overrightarrow{\mathrm{r}}_1$ and a final position $\overrightarrow{\mathrm{r}}_2$, we integrate both sides :
$$ \int\limits_{\mathrm{V}_1}^{\mathrm{V}_2} \mathrm{dV}=-\int\limits_{\overrightarrow{\mathrm{r}}_1}^{\overrightarrow{\mathrm{r}}_2} \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $$
$$\Rightarrow $$ $$ \mathrm{V}_2-\mathrm{V}_1=-\int\limits_{\overrightarrow{\mathrm{r}}_1}^{\overrightarrow{\mathrm{r}}_2} \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $$
The electric field is given as: $\overrightarrow{\mathrm{E}}=\mathrm{Ax} \hat{\imath}+\mathrm{By} \hat{\jmath}$
The general position displacement vector is: $\mathrm{d} \overrightarrow{\mathrm{r}}=\mathrm{dxi}+\mathrm{dy} \hat{\mathrm{j}}+\mathrm{dz} \hat{\mathrm{k}}$
The initial point (Point 1) is the origin: $\left(x_1, y_1\right)=(0,0)$. Let the potential here be $V_0$.
The final point (Point 2) is: $\left(x_2, y_2\right)=(10,20)$. The potential here is $\mathrm{V}=500 \mathrm{~V}$.
$$ V_2-V_1=-\int\limits_{(0,0)}^{(10,20)}(A x \hat{\imath}+B y \hat{\jmath}) \cdot(d x \hat{\imath}+d y \hat{\jmath}+d z \hat{k}) $$
$$\Rightarrow $$ $\mathrm{V}_2-\mathrm{V}_1=-\int\limits_{(0,0)}^{(10,20)} \mathrm{Axdx}+\mathrm{Bydy}$
$$\Rightarrow $$ $V_2-V_1=-\left[\int\limits_0^{10} A x d x+\int\limits_0^{20} B y d y\right]$
$$\Rightarrow $$ $\mathrm{V}_2-\mathrm{V}_1=-\left[\left[\mathrm{A}\left(\frac{\mathrm{x}^2}{2}\right)\right]_0^{10}+\left[\mathrm{B}\left(\frac{\mathrm{y}^2}{2}\right)\right]_0^{20}\right]$
$$\Rightarrow $$ $\mathrm{V}_2-\mathrm{V}_1=-\left[\mathrm{A}\left(\frac{10^2}{2}-\frac{0^2}{2}\right)+\mathrm{B}\left(\frac{20^2}{2}-\frac{0^2}{2}\right)\right]$
$$\Rightarrow $$ $\mathrm{V}_2-\mathrm{V}_1=-\left[\mathrm{A}\left(\frac{100}{2}\right)+\mathrm{B}\left(\frac{400}{2}\right)\right]$
$$\Rightarrow $$$$ V_2-V_1=-[50 A+200 B] $$
We are given : $\mathrm{A}=10, \mathrm{~B}=5, \mathrm{~V}_2=500 \mathrm{~V}$
$$ 500-V_0=-[50(10)+200(5)] $$
$$\Rightarrow $$ $500-V_0=-[500+1000]$
$$\Rightarrow $$ $500-V_0=-1500$
$$\Rightarrow $$ $\mathrm{V}_0=500+1500$
$$\Rightarrow $$ $\mathrm{V}_0=2000 \mathrm{~V}$

Answer

1000

Answer

0

Answer

2000

Answer

500

Electrostats & Capacitance

Electric field in a region is given by \vec{E}=A x \hat{i}+B y \hat{j}, where A=10 \mathrm{~V} / \mathrm{m}^2 and B=5 \mathrm{~V} / \mathrm{m}^2. If the electric potential at a point (10,20) is 500 V , then the electric potential at origin is \_\_\_\_ V.

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