Determine the magnitude of the electric field at point P in the...

For configuration (a):

The electric field at point P due to a point charge $q$ at distance $d$ is:

$E_a = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{d^2}$

For configuration (b):

The system is an electric dipole with charges $-q$ and $+q$ separated by $2a$. Point P is on the equatorial line at distance $d$.

The electric field for $d \gg a$ is:

$E_b = \dfrac{1}{4\pi\varepsilon_0} \dfrac{p}{d^3}$

where $p = 2qa$ is the dipole moment.

For configuration (c):

The system has three charges: $-q$ at $x = -a$, $+q$ at $x = 0$, and $+q$ at $x = a$. Point P is at $(0, d)$.

The field due to the central charge $+q$ is:

$\vec{E}_1 = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{d^2} \hat{j}$

The field due to the charge $-q$ at $(-a, 0)$ is:

$\vec{E}_2 = \dfrac{1}{4\pi\varepsilon_0} \dfrac{-q}{(a^2 + d^2)^{3/2}} (a\hat{i} + d\hat{j})$

The field due to the charge $+q$ at $(a, 0)$ is:

$\vec{E}_3 = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{(a^2 + d^2)^{3/2}} (-a\hat{i} + d\hat{j})$

The sum of the fields from the outer charges is:

$\vec{E}_2 + \vec{E}_3 = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{(a^2 + d^2)^{3/2}} (-2a\hat{i}) = -\dfrac{1}{4\pi\varepsilon_0} \dfrac{p}{(a^2 + d^2)^{3/2}} \hat{i}$

For $d \gg a$, this approximates to:

$\vec{E}_{23} \approx -\dfrac{1}{4\pi\varepsilon_0} \dfrac{p}{d^3} \hat{i}$

The total electric field at P is:

$\vec{E}_c = \vec{E}_1 + \vec{E}_{23} \approx -\dfrac{1}{4\pi\varepsilon_0} \dfrac{p}{d^3} \hat{i} + \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{d^2} \hat{j}$

The magnitude of the total electric field is:

$E_c = \sqrt{\left( \dfrac{p}{4\pi\varepsilon_0 d^3} \right)^2 + \left( \dfrac{q}{4\pi\varepsilon_0 d^2} \right)^2} = \dfrac{1}{4\pi\varepsilon_0 d^3} \sqrt{p^2 + q^2 d^2}$

Answer: (a) $\dfrac{q}{4\pi\varepsilon_0 d^2}$
(b) $\dfrac{p}{4\pi\varepsilon_0 d^3}$
(c) $\dfrac{1}{4\pi\varepsilon_0 d^3}\sqrt{q^2 d^2 + p^2}$