Charges are placed on the vertices of a squar | vedclass

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

Question

(D) The electric potential $V$ at the centre of the square is the algebraic sum of potentials due to individual charges: $V = k \sum \frac{q_i}{r}$. S

Vedclass · Accepted Answer

$\overrightarrow{E}$ changes,$V$ remains unchanged

Vedclass · Answer

(D) The electric potential $V$ at the centre of the square is the algebraic sum of potentials due to individual charges: $V = k \sum \frac{q_i}{r}$. Since the distance $r$ from each vertex to the centre is the same,$V = \frac{k}{r} (q_A + q_B + q_C + q_D)$. Interchanging charges does not change the sum $(q_A + q_B + q_C + q_D)$,so $V$ remains unchanged.The electric field $\overrightarrow{E}$ at the centre is a vector sum: $\overrightarrow{E} = \sum \overrightarrow{E_i}$. Initially,the field vectors from $A$ and $C$ (both $q$) and $B$ and $D$ (both $-q$) result in a net field. When charges are interchanged,the positions of the charges change relative to the centre,which alters the direction and magnitude of the individual field vectors $\overrightarrow{E_i}$. Thus,the resultant vector $\overrightarrow{E}$ changes.

Charges are placed on the vertices of a square as shown. Let $\overrightarrow{E}$ be the electric field and $V$ be the electric potential at the centre. If the charges on $A$ and $B$ are interchanged with those on $D$ and $C$ respectively,then:

Similar Questions

Two wires $A$ and $B$ of the same material,having radii in the ratio $1: 2$,carry currents in the ratio $4: 1$. The ratio of drift speed of electrons in $A$ and $B$ is

Restriction enzymes were discovered by

If $\int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx = A \tan^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}\right) + C$,where $C$ is a constant,then $A$ equals to

The major product in the following reaction is:

The contraction of muscle of shortest duration is seen in

Vedclass Test Series

Exam Paper Generator

Online Exam Module