Equal charges q are placed at the four corner | vedclass

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

Question

(C) Let $k = \frac{1}{4\pi \varepsilon_0}$. The charge at $B$ experiences forces from charges at $A, C,$ and $D$.$1$. Force due to charge at $A$: $F_A

Vedclass · Accepted Answer

$\left( \frac{1 + 2\sqrt{2}}{2} \right) \frac{q^2}{4\pi \varepsilon_0 a^2}$

Vedclass · Answer

(C) Let $k = \frac{1}{4\pi \varepsilon_0}$. The charge at $B$ experiences forces from charges at $A, C,$ and $D$.$1$. Force due to charge at $A$: $F_A = \frac{kq^2}{a^2}$ (directed along $AB$ extended).$2$. Force due to charge at $C$: $F_C = \frac{kq^2}{a^2}$ (directed along $CB$ extended).$3$. Force due to charge at $D$: $F_D = \frac{kq^2}{(a\sqrt{2})^2} = \frac{kq^2}{2a^2}$ (directed along $DB$ extended).Since $F_A$ and $F_C$ are perpendicular,their resultant $F_{AC} = \sqrt{F_A^2 + F_C^2} = \sqrt{\left(\frac{kq^2}{a^2}\right)^2 + \left(\frac{kq^2}{a^2}\right)^2} = \sqrt{2} \frac{kq^2}{a^2}$.This resultant $F_{AC}$ acts in the same direction as $F_D$ (along the diagonal $DB$).Therefore,the net force $F_{net} = F_{AC} + F_D = \sqrt{2} \frac{kq^2}{a^2} + \frac{kq^2}{2a^2} = \frac{kq^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{kq^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$.Substituting $k = \frac{1}{4\pi \varepsilon_0}$,we get $F_{net} = \left( \frac{1 + 2\sqrt{2}}{2} \right) \frac{q^2}{4\pi \varepsilon_0 a^2}$.

Equal charges $q$ are placed at the four corners $A, B, C, D$ of a square of side length $a$. The magnitude of the net electrostatic force on the charge at $B$ is:

Similar Questions

Two point charges of $1 \, \mu\text{C}$ and $5 \, \mu\text{C}$ are placed at a distance of $4 \, \text{cm}$ apart. The ratio of the forces exerted by them on each other will be:

Two point charges $A$ and $B$,having charges $+Q$ and $-Q$ respectively,are placed at a certain distance apart and the force acting between them is $F$. If $25 \%$ of the charge of $A$ is transferred to $B$,then the force between the charges becomes

$A$ charged particle of mass $m_{1}$ and charge $q_{1}$ is revolving in a circle of radius $r$. Another charged particle of charge $q_{2}$ and mass $m_{2}$ is situated at the centre of the circle. If the velocity and time period of the revolving particle are $v$ and $T$ respectively, then:

Two charges $q_1$ and $q_2$ are placed in vacuum at a distance $d$ and the force acting between them is $F$. If a medium of dielectric constant $4$ is introduced around them,the force now will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module