Question

Two point charges and are located at points (a, 0, 0) and (0, b, 0) respectively. Find the electric field due to both these charges at point (0, 0, c).

Answer 1

Hint : The electric field at any point can be given as the sum of the electric field due to all the charges near the point. We must consider the direction of the displacement vector for calculating the electric field at that point.

Formula used: In this solution we will be using the following formula;
 $\Rightarrow E = k\dfrac{q}{{{r^2}}}\hat r $ where $ E $ is the electric field due to one charge, $ q $ is the charge, $ r $ is the distance from the point of interest, $ k $ is a constant of proportionality (which can be given as $ k = \dfrac{1}{{4\pi {\varepsilon _0}}} $ where $ {\varepsilon _0} $ is the permittivity of free space) and $ \hat r $ is the unit vector in the radial direction.
 $\Rightarrow \hat r = \dfrac{{\vec r}}{r} $ , where $ \vec r $ is the displacement vector.

Complete step by step answer
To calculate the electric field at the point (0, 0, c), we have to find the sum of the electric fields due to the charges at (a, 0, 0) and (0, b, 0). Then electric field is given by
 $\Rightarrow E = k\dfrac{{{q_1}}}{{{r_1}^2}}\hat r + k\dfrac{{{q_2}}}{{{r_2}^2}}\hat r $ where $ q $ are the charges, $ r $ is the distance from the point of interest, $ k $ is a constant of proportionality, and $ \hat r $ is the unit vector in the radial direction.
And $ \hat r $ is given by
 $\Rightarrow \hat r = \dfrac{{\vec r}}{r} $ where $ \vec r $ is the displacement vector. Hence, by substitution
 $\Rightarrow E = k\dfrac{{{q_1}}}{{{r^3}}}{\vec r_1} + k\dfrac{{{q_2}}}{{{r^3}}}{\vec r_2} $ .
 $\Rightarrow {\vec r_1} = - a\hat i + c\hat j $ and
 $\Rightarrow {\vec r_2} = - b\hat i + c\hat j $
This is because to locate (0, 0, c) from both axis, we must be pointing in the opposite direction of the axis.
Then, $ {r_1} = \sqrt {{a^2} + {c^2}} $ and $ {r_2} = \sqrt {{b^2} + {c^2}} $ , inserting into the equation, we have
 $\Rightarrow E = \dfrac{{k{q_1}\left( { - a\hat i + c\hat j} \right)}}{{{{\left( {\sqrt {{a^2} + {c^2}} } \right)}^3}}} + \dfrac{{k{q_2}\left( { - b\hat i + c\hat j} \right)}}{{{{\left( {\sqrt {{b^2} + {c^2}} } \right)}^3}}} $
The constant of proportionality $ k $ , can be given as
 $\Rightarrow k = \dfrac{1}{{4\pi {\varepsilon _0}}} $ where $ {\varepsilon _0} $ is the permittivity of free space.
Hence the electric field is
 $\Rightarrow E = \dfrac{{k{q_1}\left( { - a\hat i + c\hat j} \right)}}{{{{\left( {{a^2} + {c^2}} \right)}^{\dfrac{3}{2}}}}} + \dfrac{{k{q_2}\left( { - b\hat i + c\hat j} \right)}}{{{{\left( {{b^2} + {c^2}} \right)}^{\dfrac{3}{2}}}}} $.

Note
For clarity, the displacement vector can be mathematically given as
 $\Rightarrow r = ({x_2},{y_2},{z_2}) - ({x_1},{y_2},{z_2}) $
Hence, for $ {\vec r_1} $ we have $ {\vec r_1} = (0,0,c) - (a,0,0) = ( - a,0,c) $
Which can be written in vector form as $ - a\hat i + c\hat j $
Similarly for $ {\vec r_2} $ , we have $ {\vec r_2} = (0,0,c) - (0,b,0) = (0, - b,0) $
Which can be written in vector form as $ - b\hat i + c\hat j $.