Question 116 of 311

Four point charges $-q,+q,+q$ and $-q$ are placed on y-axis at $y=-2d,y=-d,$ and $y=+2d,$ respectively. The magnitude of the electric field $E$ at a point on the x-axis at $x=D,$ with $D≫d,$ will behave as:

A$E\propto \dfrac{1}{{D}^{4}}$B$E\propto \dfrac{1}{D}$C$E\propto \dfrac{1}{{D}^{3}}$D$E\propto \dfrac{1}{{D}^{2}}$

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Answer: (A) $E\propto \dfrac{1}{{D}^{4}}$

The electric field at $P$ is superposition of the fields due to all four charges. The electric field due to the two $+q$ charges is

${E}_{+q}=\dfrac{Kq}{{D}^{2}+{d}^{2}}{\cos ⁡\theta }_{1}\times 2$
${E}_{-q}=\dfrac{Kq}{{D}^{2}+4{d}^{2}}{\cos ⁡\theta }_{2}\times 2$
The field due to $+q$ is towards right and the field due to $-q$ is towards left. Also ${\theta }_{1}\approx 0$ and ${\theta }_{2}\approx 0$ .
$∴{E}_{net}={E}_{+q}-{E}_{-q}\approx \dfrac{Kq}{{D}^{2}+{d}^{2}}-\dfrac{Kq}{{D}^{2}+4{d}^{2}}$
$=\dfrac{Kq3{d}^{2}}{\left({D}^{2}+{d}^{2}\right)\left({D}^{2}+4{d}^{2}\right)}\approx \dfrac{3Kq{d}^{2}}{{D}^{4}}$
$E\propto \dfrac{1}{{D}^{4}}$