Spherical symmetry. a. A non-conducting sphere of radius R has charge Q distribu

Question

Spherical symmetry.

a. A non-conducting sphere of radius R has charge Q distributed uniformly

through its volume. Find the magnitude of E at distance r from the sphere’s center for r > R . Ans: E = kQ/r2 .

b. Repeatfor r<R.Ans: E=kQr/R3.

c. Let this sphere be surrounded by a concentric conducting shell, with zero

net charge, inner radius R1 , and outer radius R2 (both > R ). Find the charge on the inner surface at r = R1 . Ans: –Q.

d. For this combination make a careful plot of E vs. r.

A non-conducting sphere of radius R has charge Q distributed uniformly through its volume. Find the magnitude of E at distance r from the sphere's center for r >R. Ans: E kQ/r2 b. Repeat for r R. Ans: E kQr/R3 Let this sphere be surrounded by a concentric conducting shell, with zero net charge, inner radius Ri, and outer radius R2 (both > R). Find the charge on the inner surface at r - C. For this combination make a careful plot of E vs. r.

Explanation / Answer

a] By Gauss law, taking spherical Gaussian surface of concentric center and radius r,

E. area = Qenclosed/e0

E 4 pi r^2 =  Q/e0

E =  Q/4pi e0 r^2

= kQ/r^2

b] Again,

E. area = Qenclosed/e0

E 4 pi r^2 =  Q*(r^3/R^3)/e0

E =  Qr/4pi e0 R^3

= kQr/R^3

c] Let take spherical Gaussian surface of concentric center and radius r which is greater than R1 but smaller than R2,let charge on inner surface be q, we also know that electric field is zero inside conductor

E. area = Qenclosed/e0

0* 4 pi r^2 = [Q+q] /e0

q+Q = 0

q = -Q

d] For ploting, E increases linearly with r for r<R,

then decreases quadratically with r as E = kQ/r^2 for R < r<R1

then E=0 for r > R1

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