Question

Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origin, another at (2 cm, 0, 0) and the third at (0, 2 cm, 0). Find the force acting on the charge located at the origin.

Answer 1

`\vec F_(A) = -67500\,N\cdot (i + j)`

Step-by-step explanation:

The position of each point are the following:

`A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)`

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

`\vec F_(A) = \vec F_(AB) + \vec F_(AC)`

`\vec F_(A) = -(k\cdot q^(2))/(r_(AB)^(2))\cdot i - (k\cdot q^(2))/(r_(AC)^(2))\cdot j`

`\vec F_(A) = - (k\cdot q^(2))/(r^(2)) \cdot (i + j)`

`\vec F_(A) = -((9 * 10^(9)\,(N\cdot m^(2))/(C^(2)) )\cdot (3* 10^(-9)\,C))/((0.02\,m)^(2))\cdot (i + j)`

`\vec F_(A) = -67500\,N\cdot (i + j)`