Answer:
54 n C
Explanation:
given,
Electric field strength = 60,000 N/m
diameter of the sphere = 10 cm
distance of the electric field = 4 cm
distance of the point form the center of ball
r = 10/2 + 4
r = 9 cm = 0.09 m
the electric field strength
[tex]E = \dfrac{kq}{r^2}[/tex]
[tex]60000 = \dfrac{9\times 10^9\times q}{0.09^2}[/tex]
[tex]q = \dfrac{60000\times 0.09^2}{9\times 10^9}[/tex]
q = 54 x 10⁻⁹ C
q = 54 n C
the charge on the ball is equal to 54 n C
The magnitude of charge in the ball is 54 nC.
Given data:
The strength of Electric field is, E = 60,000 N/C.
The diameter of ball is, d = 10 cm = 0.1 m.
The distance is, r = 4.0 cm = 0.04 cm.
The region where the electric force has its significance is known as electric field strength. And the expression for electric field strength is given as,
[tex]E = \dfrac{kq}{(r+d/2)^{2}}[/tex]
Here, k is the Coulomb's constant and q is the magnitude of charge.
Solving as,
[tex]60,000 = \dfrac{9 \times 10^{9} \times q}{(0.04+0.1/2)^{2}}\\\\q =54 \times 10^{-9} \;\rm C\\\\q = 54 \;\rm nC[/tex]
Thus, we can conclude that the magnitude of charge in the ball is 54 nC.
Learn more about the electric field strength here:
https://brainly.com/question/15170044
