Question

Two point charges are placed at two of the corners of a triangle as shown in the figure.Find the magnitude and the direction of the electric field at the third corner of the triangle.
point 1 :100 με
distance from point 1 to 3 :100 cm
point 2 : +150 pc
distance from point 3 to 2 : 200 cm-
point 3 :?

Answer 1

To find the magnitude and direction of the electric field at point 3, we can use Coulomb's law.

Coulomb's law states that the electric field created by a point charge is given by:

E = k * Q / r^2

Where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 N m^2 / C^2), Q is the charge, and r is the distance from the charge.

First, let's find the electric field created by point charge 1 at point 3:

E1 = k * Q1 / r1^2
= (8.99 x 10^9 N m^2 / C^2) * (100 * 10^-6 C) / (100 * 10^-2 m)^2
= 8.99 x 10^6 N/C

The electric field created by point charge 1 is 8.99 x 10^6 N/C directed towards point 3.

Next, let's find the electric field created by point charge 2 at point 3:

E2 = k * Q2 / r2^2
= (8.99 x 10^9 N m^2 / C^2) * (150 * 10^-12 C) / (200 * 10^-2 m)^2
= 1.124 x 10^5 N/C

The electric field created by point charge 2 is 1.124 x 10^5 N/C directed towards point 3.

To find the total electric field at point 3, we can add the two electric fields:

E_total = E1 + E2
= 8.99 x 10^6 N/C + 1.124 x 10^5 N/C
= 9.112 x 10^6 N/C

The magnitude of the electric field at point 3 is 9.112 x 10^6 N/C. Since both charges are pointing towards point 3, the direction of the electric field at point 3 is also towards point 3.