Question

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge. Find the magnitude of the electric field at the center of the square. (k=1/4πϵ0=8.99×109 N.m^2/C^2)

Answer 1

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$

$E =E_2 + E_4$

$E = 2 E_2$

[$E_2 =\frac{2\times k \times q}{r^2}$

[$r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m$]

plugging all value

$E = 2 E_2$

$E = 2 E_2 =\frac{2\times k \times q}{r^2}$

$E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}$

E = 440816.32 N/C

Answer 2

Answer:

$E=4.32\times 10^{5}\ N.C^{-1}$

Explanation:

Given:

• charges at the each of the three corner of a square, $q_1=q_2=q_3=+3\times 10^{-6}\ C$

• side of the square, $a=0.5\ m$

• charge at the remaining corner of the square, $q_4=-3\times 10^{-6}\ C$

Distance of the center of the square from each of the vertex of square:

Using Pythagoras theorem:

$d^2+d^2=a^2$

$2d^2=0.5^2$

$d=0.3536\ m$

As there two equal like charges at an equal distance on the opposite ends of a diagonal so they will cancel out the effect of field due to each other.

Electric field at the center of the square due to $q_4$:

$E_4=\frac{1}{4\pi.\epsilon_0} \times \frac{q_4}{d^2}$

$E_4=9\times 10^9\times\frac{3\times 10^{-6}}{0.125}$

$E_4=216\times10^{3}\ N.C^{-1}$

The charge on the opposite vertex will have the equal effect in the same direction which is towards the charge $q_4$. (refer the attached schematic)

So, the net electric field at the center:

$E=2\times E_4$

$E=2\times 216\times 10^3$

$E=4.32\times 10^{5}\ N.C^{-1}$