Question 137: Electrostatics & Capacitance

Question

The electric potential as a function of $x, y$ is given by $V=5\left(x^2-y^2\right) V$. The electric field at a point $(2,3) \mathrm{m}$ is $\_\_\_\_$ $\mathrm{V} / \mathrm{m}$.

Accepted Answer

The electric field $\vec{E}$ is the negative gradient of the electric potential $V$ :
$$ \overrightarrow{\mathrm{E}}=-
abla \mathrm{V}=-\left(\frac{\partial \mathrm{V}}{\partial \mathrm{x}} \hat{\mathrm{i}}+\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \hat{\mathrm{j}}+\frac{\partial \mathrm{V}}{\partial \mathrm{z}} \hat{\mathrm{k}}ight) $$
The electric potential is given as $V=5\left(x^2-y^2ight)=5 x^2-5 y^2$.
The x -component of electric field is,
$$ E_x=-\frac{\partial V}{\partial x}=-\frac{\partial}{\partial x}\left(5 x^2-5 y^2ight)=-(10 x)=-10 x $$
The y -component of electric field is,
$$ E_y=-\frac{\partial V}{\partial y}=-\frac{\partial}{\partial y}\left(5 x^2-5 y^2ight)=(10 y)=10 x $$
At Point $(2,3)$
$$ \mathrm{E}_{\mathrm{x}}=-10(2)=-20 \mathrm{~V} / \mathrm{m} $$
$$\Rightarrow $$ $$ \mathrm{E}_{\mathrm{y}}=10(3)=30 \mathrm{~V} / \mathrm{m} $$
So, the resulting electric field vector is :
$$ \overrightarrow{\mathrm{E}}=-20 \hat{\imath}+30 \hat{\jmath} $$
Therefore, the correct option is (A).

Answer

$(-20 \hat{i}+30 \hat{j})$

Answer

$(20 \hat{i}-30 \hat{j})$

Answer

$(20 \hat{i}+45 \hat{j})$

Answer

$(-4 \hat{i}+6 \hat{j})$

Electrostatics & Capacitance

The electric potential as a function of x, y is given by V=5\left(x^2-y^2\right) V. The electric field at a point (2,3) \mathrm{m} is \_\_\_\_ \mathrm{V} / \mathrm{m}.

Choose the correct answer: