Respuesta :
Answer and Solution:
As per the question:
Uniform Charge distribution on the hollow sphere, q = - 9 nC = [tex]9\times {- 9} C[/tex]
Separation distance between the center of the two balls, d = 7 cm = 0.07 m
Radius of the metal ball, R = 1.5 cm = 0.015 m
Radius of the hollow ball, R' = 2 cm = 0.02 m
Now,
(a) The hollow ball has a uniformly distributed negative charge of - 9 nC so it will induce + 9 nC of charge on the solid ball near it.
The surface of the ball closer to the negatively charged hollow ball will get positive charge due to the force of attraction between the positive and negative charges and the negative charge of the metal ball are repelled to the farther surface of the ball.
(b) The electric field at the center of the metal ball is given by:
[tex]\vec{E} = k\frac{q^{2}}{d}[/tex]
[tex]\vec{E} = k\frac{q^{2}}{R}[/tex]
Since, the distance from the surface to the center of the metal ball is its radius.
where
k = electrostatic constant = [tex]9\times 10^{9} Nm^{2}C^{- 2}[/tex]
Thus
[tex]\vec{E} = 9\times 10^{9}\times \frac{(9\times {- 9})^{2}}{0.015} = 5400 N/C[/tex]
The electric field at the center of the metal ball due just to the charges on its surface is 5,400 N/C.
Given to us,
Radius of the metal ball = 1.5 cm = 0.015 m
Radius of the hollow ball = 2 cm = 0.02 m
Uniformly distributed charge, q = -9 nC = 9 x 10⁻⁹C
Distance between the centers of the ball, d = 7 cm = 0.07 m
Which edge of the metal ball will have an excess of positive charges?
we know that the hollow ball has a uniformly distributed charge of -9nC, therefore it will induce +9 nC on the solid ball.
Since the surface of the hollow ball is negatively charged it will try to attract the positive charge, which will lead to the repulsion of the surface of the ball further.
What is the electric field at the center of the metal ball due just to the charges on its surface?
We know that the electric field at the center of the metal ball is given by,
[tex]\overrightarrow E = k\dfrac{q^2}{d}\\\\\overrightarrow E = k\dfrac{q^2}{R}[/tex]
Substitute the values,
[tex]\overrightarrow E = (9\times 10^{9})\dfrac{(9\times 10^{-9})^2}{0.015}\\\\\overrightarrow E = 5400 \rm\ N/C[/tex]
Hence, the electric field at the center of the metal ball due just to the charges on its surface is 5,400 N/C.
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