Question

Electric charges q1 = q2 = +6 µC are at the bottom vertices of an equilateral triangle having 40 cm side.

Calculate the magnitude and direction of the electric fields due to q1 and q2 at the top vertex of the triangle.

Calculate the magnitude and direction of the electric force on charge q = -5 µC placed on the top vertex of the triangle.

Answer 1

here,

q1 = q2 = 6 uC = 6 * 10^-6 C

a = 40 cm = 0.4 m

the magnitude of electric feild due to each charge , E1 = K * q1 /a^2

E1 = 9 * 10^9 * 6 * 10^-6 /0.4^2

E1 = 0.338 N

as both equal charges are at two corners of equilateral triangle

and we have to find out the electric feild at the third corner ,

the horizontal component of electric feild cancels out

so, the direction of feild is UPwards

the magnitude of frild , E = 2 * E1 * cos(30)

E = 2 * 0.338 * cos(30)

E = 0.58 N/C

q = - 5 uC = - 5 * 10^-6 C

the magnitude of force , F = q * E

F = 0.58 * (5 * 10^-6 ) N = 2.92 * 10^-6 N

as the charge is negative

the direction of force is downwards