Question

Pls answer with explanation

Answer 1

Force experienced at B by charges at A and C will be same since they're equidistant from B.

Distance between A and B is equal to length of square 'a', same as the distance between B and C

• $\displaystyle\sf{AB=BC=a}$

Then the force due to A, or C, is,

$\displaystyle\longrightarrow\sf{F_A=F_C=\dfrac {1}{4\pi\epsilon_0}\,\dfrac {q^2}{a^2}}$

But we need their resultant.

Since $\displaystyle\sf{F_A}$ and $\displaystyle\sf{F_C}$ are perpendicular to each other,

$\displaystyle\longrightarrow\sf{F_A+F_C=\sqrt{\left (\dfrac {1}{4\pi\epsilon_0}\,\dfrac {q^2}{a^2}\right)^2+\left (\dfrac {1}{4\pi\epsilon_0}\,\dfrac {q^2}{a^2}\right)^2}}$

$\displaystyle\longrightarrow\sf{F_A+F_C=\dfrac {\sqrt2}{4\pi\epsilon_0}\,\dfrac {q^2}{a^2}}$

Distance between B and D is equal to length of diagonal of square, i.e., a√2.

• $\displaystyle\sf{BD=a\sqrt2}$

Then force due to D is,

$\displaystyle\longrightarrow\sf{F_D=\dfrac {1}{4\pi\epsilon_0}\,\dfrac {q^2}{(a\sqrt2)^2}}$

$\displaystyle\longrightarrow\sf{F_D=\dfrac {1}{4\pi\epsilon_0}\,\dfrac {q^2}{2a^2}}$

Hence magnitude of net electric force acting on B is,

$\displaystyle\longrightarrow\sf{F=F_A+F_C+F_D}$

$\displaystyle\longrightarrow\sf{F=\dfrac {1}{4\pi\epsilon_0}\,\dfrac {q^2}{2a^2}+\dfrac {\sqrt2}{4\pi\epsilon_0}\,\dfrac {q^2}{a^2}}$

$\displaystyle\longrightarrow\underline {\underline {\sf{F=\left (\dfrac {1+2\sqrt2}{2}\right)\dfrac {q^2}{4\pi\epsilon_0a^2}}}}$

Hence (c) is the answer.