Question

Four particles carrying charges + 2q, − 2q, + 4q and − 4q (with q = 1.0 nC) are kept at the vertices of a square of side 6.0 cm. Determine the net electric field due to these charged particles at the centre of the square. What is the electrostatic force on a particle carrying positive charge of 1.0 nC placed at the centre of the square?

Answer 1

please specify how the charges are placed on the vertices of the square.

Answer 2

Answer: The net electric field is $E =1.41\times10^{4}\ C/m^2$ and  force is the $F = 0.141\mu N$.

Explanation:

Given that,

Charge of first particle = +2 q

Charge of second particle = -2 q

Charge of third particle = +4 q

Charge of fourth particle = -4 q

Side of square = 6.0 m

We know that,

The electric field produced by the charge +2 q

$E_{1}= k\dfrac{2q}{(\dfrac{d}{2})^2}$

The electric field produced by the charge -2 q

$E_{2}= k\dfrac{-2q}{(\dfrac{d}{2})^2}$

The electric field produced by the charge +4 q

$E_{3}= k\dfrac{4q}{(\dfrac{d}{2})^2}$

The electric field produced by the charge -4 q

$E_{4}= k\dfrac{-4q}{(\dfrac{d}{2})^2}$

Now, the resultant electric field is between $q_{3}$ and $q_{1}$

$E_{5}= E_{3}-E_{1}$

$E _{5}= k\dfrac{4q}{(\dfrac{d}{2})^2}-k\dfrac{2q}{(\dfrac{d}{2})^2}$

$E_{5}= k\dfrac{2q}{(\dfrac{d}{2})^2}$

$E_{5}=1.0\times10^{-4}\ C/m^2$

The resultant electric field is between $q_{4}$ and $q_{2}$

$E _{6}= k\dfrac{-4q}{(\dfrac{d}{2})^2}-k\dfrac{-2q}{(\dfrac{d}{2})^2}$

$E_{6}= k\dfrac{-2q}{(\dfrac{d}{2})^2}$

$E_{6}=-1.0\times10^{-4} C/m^2$

The resultant electric field is

$E=\sqrt{E_{5}^2+E_{6}^2}$

$E = \sqrt{(1.0\times10^{-4})^2+(1.0\times10^{-4})^{2}}$

$E =1.41\times10^{4}\ C/m^2$

The electrostatic force is the product of the electric field and charge.

$F = qE$

$F = 1.0\times10^{-9}\times1.41\times10^{4}$

$F = 1.41\times10^{-5}N$

$F = 0.141\mu N$

Hence, The net electric field is $E =1.41\times10^{4}\ C/m^2$ and  force is the $F = 0.141\mu N$.