The electric charges +2q, +2q, -2q and -2q ar | vedclass

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

Question

(B) The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.Point $A$ is at

Vedclass · Accepted Answer

$\frac{q}{\pi \epsilon_0 L}\left[1-\frac{1}{\sqrt{5}}\right]$

Vedclass · Answer

(B) The electric potential $V$ at a point due to a charge $q$ at a distance $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.Point $A$ is at the midpoint of the side with charges $+2q$ and $+2q$. The distance of $A$ from each of these charges is $L$.The distance of $A$ from each of the charges $-2q$ at the opposite corners can be found using the Pythagorean theorem. The horizontal distance is $2L$ and the vertical distance is $L$, so the distance $r = \sqrt{(2L)^2 + L^2} = \sqrt{5L^2} = L\sqrt{5}$.The total potential at $A$ is the sum of potentials due to all four charges:$V_A = \frac{1}{4\pi\epsilon_0} \left[ \frac{2q}{L} + \frac{2q}{L} + \frac{-2q}{L\sqrt{5}} + \frac{-2q}{L\sqrt{5}} \right]$$V_A = \frac{1}{4\pi\epsilon_0} \left[ \frac{4q}{L} - \frac{4q}{L\sqrt{5}} \right]$$V_A = \frac{4q}{4\pi\epsilon_0 L} \left[ 1 - \frac{1}{\sqrt{5}} \right] = \frac{q}{\pi\epsilon_0 L} \left[ 1 - \frac{1}{\sqrt{5}} \right]$.

The electric charges $+2q$, $+2q$, $-2q$ and $-2q$ are placed at the corners of a square of side $2L$ as shown in the figure. The electric potential at point $A$, midway between the two charges $+2q$ and $+2q$, is $(\epsilon_0 = \text{permittivity of free space})$

Similar Questions

$A$ charge $2\,\mu C$ is taken from infinity to a point in an electric field,without changing its velocity. If the work done against the electric forces is $20\,\mu J$,then the potential at that point will be.....$V$

Two charges of $4\,\mu C$ each are placed at the corners $A$ and $B$ of an equilateral triangle of side length $0.2\, m$ in air. Find the electric potential at $C$. $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\text{N m}^2/\text{C}^2} \right]$

The total charge on a uniformly charged spherical shell having radius $R$ is $Q$. Then electric potential at a distance $r = R/2$ from the centre of the shell is . . . . . . .

In a uniform electric field,the potential is $10 \ V$ at the origin of coordinates,and $8 \ V$ at each of the points $(1, 0, 0), (0, 1, 0)$ and $(0, 0, 1)$. The potential at the point $(1, 1, 1)$ will be....$V$

Vedclass Test Series

Exam Paper Generator

Online Exam Module