Question

A charge +40x10-6C is 0.04 m above an unknown charge q. The electric intensity of a point at 0.01 m above a charge +40x10-6C is of 5x109N/C and is directed upward. Find the magnitude and sign of the unknown charge.

Answer 1

Answer:

q1 = 40mC = 40*10⁻³ C

q2 =20mC = 20*10⁻³ C

E = 5*10⁹N/C

K = 9*10⁹N*m²/C

r = 0,01 m

¿Cuál es la magnitud y el signo de la carga desconocida?

Para determinar la magnitud de la carga desconocida, aplicaremos la ley de Coulomb para determinar la fuerza aplicada a las cargas:

F = K*q1*q2/r²

F = 9*10⁹N*m²/C*40*10⁻³ C* 20*10⁻³ C/ (0,01m)²

F = 112.000.000N = 11,25*10⁷ N

E = F/q3

q3 = F/E

q3 = 11,25*10⁷ N/ 5*10⁹N/C

q3 = 2,25*10¹⁶

Answer 2

Answer:

q3 = 2,25*10¹⁶

hope it helps u !!