Three charges,each +q,are placed at the corne | vedclass

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

Question

(D) Given that $AC = BC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively.Therefore,$AE = EC = a$ and $BD = DC = a$.In $\Delta ADC$,b

Vedclass · Accepted Answer

zero

Vedclass · Answer

(D) Given that $AC = BC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$ respectively.Therefore,$AE = EC = a$ and $BD = DC = a$.In $\Delta ADC$,by the Pythagorean theorem,$(AD)^2 = (AC)^2 - (DC)^2 = (2a)^2 - (a)^2 = 4a^2 - a^2 = 3a^2$,so $AD = a\sqrt{3}$.Similarly,in $\Delta BEC$,$(BE)^2 = (BC)^2 - (EC)^2 = (2a)^2 - (a)^2 = 3a^2$,so $BE = a\sqrt{3}$.The electric potential at point $D$ due to the charges at $A, B,$ and $C$ is:$V_D = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{BD} + \frac{q}{DC} + \frac{q}{AD} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{a} + \frac{1}{a} + \frac{1}{a\sqrt{3}} \right] = \frac{q}{4\pi\varepsilon_0 a} \left[ 2 + \frac{1}{\sqrt{3}} \right]$.The electric potential at point $E$ due to the charges at $A, B,$ and $C$ is:$V_E = \frac{1}{4\pi\varepsilon_0} \left[ \frac{q}{AE} + \frac{q}{EC} + \frac{q}{BE} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{a} + \frac{1}{a} + \frac{1}{a\sqrt{3}} \right] = \frac{q}{4\pi\varepsilon_0 a} \left[ 2 + \frac{1}{\sqrt{3}} \right]$.Since $V_D = V_E$,the work done $W = Q(V_E - V_D) = 0$.

Three charges,each $+q$,are placed at the corners of an isosceles triangle $ABC$ with sides $BC = AC = 2a$. $D$ and $E$ are the midpoints of $BC$ and $AC$,respectively. The work done in taking a charge $Q$ from $D$ to $E$ is

Similar Questions

Two positive charges $Q$ and $4Q$ are placed at points $A$ and $B$ respectively, where $B$ is at a distance $d$ units to the right of $A$. The total electric potential due to these charges is minimum at point $P$ on the line segment joining $A$ and $B$. What is the distance of $P$ from $A$?

$A$ charge of $5\,C$ is given a displacement of $0.5\,m$. The work done in the process is $10\,J$. The potential difference between the two points will be.......$V$

Calculate the electric potential on the axis of a disc of radius $R$ due to a charge $Q$ uniformly distributed on its surface.

Two point charges $q_1$ and $q_2$ are separated by a distance $d$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by a distance $x$? $(x < d)$ (where $\frac{1}{4 \pi \varepsilon_0} = K$ is a constant).

$1 \ eV = ...... \ J$. (Fill in the gap)

Vedclass Test Series

Exam Paper Generator

Online Exam Module