A uniform electric field of 20, N/C exists al | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electric field is given as $\vec{E} = 20\hat{i}\, N/C$.For a uniform electric field,the potential difference between two points $A$ and $B$ is

Vedclass · Accepted Answer

$-40$

Vedclass · Answer

(B) The electric field is given as $\vec{E} = 20\hat{i}\, N/C$.For a uniform electric field,the potential difference between two points $A$ and $B$ is given by the relation $\Delta V = V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.Here,the displacement vector $\vec{r}_{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} = (6 - 4)\hat{i} + (5 - 2)\hat{j} = 2\hat{i} + 3\hat{j}$.Substituting the values,we get $V_B - V_A = -(20\hat{i}) \cdot (2\hat{i} + 3\hat{j})$.$V_B - V_A = -20 	imes 2 = -40\, V$.

$A$ uniform electric field of $20\, N/C$ exists along the $x$-axis in a space. The potential difference $(V_B - V_A)$ for the points $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$.

Similar Questions

$A$ parallel plate capacitor has a potential of $20 \ kV$ and a capacitance of $2 \times 10^{-4} \ \mu F$. If the area of the plate is $0.01 \ m^2$ and the distance between the plates is $2 \ mm$,find the potential gradient.

Variation in electric potential is maximum if one goes

Electric field at a point $(x, y, z)$ is represented by $\vec E = 2x\hat i + y^2\hat j$. If the potential at $(0, 0, 0)$ is $2 \, V$,find the potential at $(1, 1, 1)$. (in $/3$)

If the electric potential is given by $V = (5x^2 + 10x - 9) \ V$,what is the electric field at $x = 1 \ m$ in $V/m$?

The potential gradient is a

Vedclass Test Series

Exam Paper Generator

Online Exam Module