a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly over its surface. find the magnitude and direction of the electric field (a) 2.2cm,(b)5.6cm,and (c)14 cm from the point charge.

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

tallinn tallinn · Answer 2 · 2022-02-03T15:49:02+01:00

A) when E is = 278925.62 N/C with direction; radially out.
B) When E is = 43048.47 N/C with direction radially out.
C) When E is = -3214.29 N/C with direction radially in.

When From Gauss' Law, also the Electric field for any spherically symmetric charge or also that charge distribution is the same as the point charge formula. Thus;

Then E = kQ/r²
After that Q is the net charge within the distance r.
Then We are given the charge Q = 15-nC and also a spherical shell of a radius 10cm

A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C

Then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²
Then E = 278925.62 N/C
Then This will be radially out since the net charge is positive.

B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C

then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²
Then E = 43048.47 N/C
After that This will be radially out since the net charge is positive.

C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Then Q = 15 nC - 22 nC
After that Q = -7 nC = -7 x 10^(-9) C
When E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²
Then E = -3214.29 N/C

Thus, This will be radially in, since the net charge is negative.

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