11. -/1 points HRW10 23.PO39.ssm. My Notes Ask Your Teacher In the figure below,

Question

11. -/1 points HRW10 23.PO39.ssm. My Notes Ask Your Teacher In the figure below, a small, nonconducting ball of mass m -1.5 mg and charge q 2.2 x 10* C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle 0- 34° with a vertical, uniformly charged nonconducting sheet (shown in cross section) Considering the gravitational force of the ball and assuming that the sheet extends far vertically and into and out of the page, calculate the surface charge density of the sheet. - 1.5 mg and charge q 2.2x10-8 c C/m2 n,q Additional Materials Section 23.5

Explanation / Answer

Given that, m = 1.5 mg = 1.5 x 10^-6 kg.

As we know that the magnitude of the electric field due to an infinite charged sheet is given by -  

E = /(2*)

where is the charge density of the sheet (charge per unit area) and is the permittivity of the medium surrounding the sheet.

Now, for a vacuum, = o = 8.854*10^-12 F/m

= 8.854*10^-12 (C^2)/(m*J)

Again, the magnitude of the electrostatic force on the ball is |F_e| = q*E = q*/(2*), where q is the charge on the ball.

This force is directed perpendicular to the plane of the charged sheet. The direction of the force depends on whether the charge on the sheet and the ball are of the same sign, which is not specified in the problem.

The magnitude of the gravitational force (assuming this setup is near the surface of the Earth) is |F_grav} = m*g, and directed downward.

Here, the ball is not moving, so the net force on the ball is zero. The force exerted on the ball by the tension in the string is therefore equal to the vector sum of the gravitational and electrostatic forces.

If you draw a force diagram, it should be clear that:

|F_e|/|F_grav| = tan(34 deg)

So:

(|q*|/(2*))/(m*g) = tan(34 deg)

|| = (2*m*g*/|q|)*tan(34 deg)

Plugging in the values for this problem, and assuming the medium surrounding the ball and sheet is a vacuum, we have:

|| = tan(10 deg) * 2*(1.5*10^-6 kg)*(9.8 m/s^2)*(8.854*10^-12 (C^2)/(m*J))/(2.2*10^-8 C)

|| =(1.18 x 10^-8 C/m^2)*tan(10 deg) = 2.08 x 10^-9 C/m^2 (Answer)

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