A 10- point charge is located at the center of a thin spherical shell of radius

Question

A 10- point charge is located at the center of a thin spherical shell of radius 8.0 carrying -20 distributed uniformly over its surface.

A. What is the magnitude of the electric field 2.0 from the point charge?
B. What is the direction of the electric field in part (A)?
C. What is the magnitude of the electric field 6.0 from the point charge?
D. What is the direction of the electric field in part (C)
E. What is the magnitude of the electric field 15 from the point charge?
F. What is the direction of the electric field in part (E)

Explanation / Answer

Before we begin, recall that the electric field of a point charge is given by:

E = k * Q / (r^2)

where k = 1/(4*pi*e0) = 9 * 10^9

Let's begin:

1) 2cm = 2 * 10 ^-2 m = .02 m. This is within the sphere, so we only worry about the charge contained within a .02 m radius. The charge contained within the radius is

Q = 10 nC = 10 * 10^-9 C

So our electric field becomes:

E = k * Q / (r^2) = 9*10^9 * 10 * 10 ^-9 / (.02)^2 = 90 / .0004 = 2.25*10^5 N/C

2) Similarly as (1), 6 cm = .06 m which is still within the sphere. So our electric field looks the same but with a different radius:

E = k * Q / (r^2) = 9*10^9 * 10 * 10 ^-9 / (.06)^2 = 90 / .0036 = 2.5*10^4 N/C

3) 15 cm = .15 m. We are now outside of the sphere. The total charge contained within the .15 m radius acts like a point charge.

So we add up the total charge and use the same formula for E above with a radius of 0.15 m. So we get:

Q = ( 10 - 20 ) * 10^ -9 C = -10*10^-9 C = -10^-8 C

E = k * Q / (r^2) = 9*10^9 * -10^-8 / (.15^2) = -90 / .0225 = -4000 N/C

Hope that helps.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.