Question

Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E midway between these two charges?

Answer 1

Answer:

The intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

$|E_{1}|=k\frac{q_{1}}{d^{2}}$

Where:

• k is Coulomb's constant
• q(1) is 20.00 μC or 20*10⁻⁶ C
• d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

$|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}$

$|E_{1}|=1.8*10^{7}\: N/C$

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

$|E_{2}|=k\frac{q_{2}}{d^{2}}$

$|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}$

$|E_{2}|=7.2*10^{6}\: N/C$

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

I hope it helps you!