Find the electric field in the plane of the page 5.0 cm to the left of P. Expres

Question

Find the electric field in the plane of the page 5.0 cm to the left of P. Express your answers using two significant figures. Enter your answers numerically separated by a comma. Assume that the positive x and y axes are directed to the right and upward respectively. Find the electric field in the plane of the page 5.0 cm directly above P . Express your answers using two significant figures. Enter your answers numerically separated by a comma. Assume that the x and y axes are directed to the right and upward respectively. Find the electric field in the plane of the page at P. Express your answers using two significant figures. Enter your answers numerically separated by a comma. Assume that the x and y axes are directed to the right and upward respectively.

Explanation / Answer

Part A)

The formula for Electric Field is E = kq/r2

We need to find the E field from each chare and add the vectors.

For the +2C charge

E = (9 X 109)(2 X 10-6)/(.025)2 = 2.88 X 107 N/C pushing toward negative x

For the -2C charge

E = (9 X 109)(-2 X 10-6)/(.075)2 =  3.20 X 106 N/C pulling toward positive  x

The net is 2.88 X 107 N/C - 3.20 X 106 N/C =

2.56 X 107 N/C directed toward the negative x

Part B

For a point 5 cm directly above P, lets first find the distance to that point from the charges

d = [(.025)2 + (.05)2]

d = .0559 m

For the +2C charge

E = (9 X 109)(2 X 10-6)/(.0559)2 =  5.76 X 106 N/C

We need to find an angle as well

tan = .05/.025

= 63.4o North of East

Therefore it has a positive x component and positive y component found by

x = (5.76 X 106 N/C)(cos 63.4) = 2.58 X 106 N/C

y = (5.76 X 106 N/C)(sin 63.4) = 5.15 X 106 N/C

For the -2C charge

E = (9 X 109)(-2 X 10-6)/(.0559)2 = 5.76 X 106 N/C

Again we need the angle, which will be the same angle, but pulling South of East

Therefore it has a positive x component and a negative y component found by

x = (5.76 X 106 N/C)(cos 63.4) = 2.58 X 106 N/C

y = (5.76 X 106 N/C)(sin 63.4) = 5.15 X 106 N/C

Therefore we find that the y components cancel each other out, but we are left with a net positive x component of

2.58 X 106 N/C + 2.58 X 106 N/C = 5.16 X 106 N/C

Part C

For the +2C charge

E = (9 X 109)(2 X 10-6)/(.025)2 = 2.88 X 107 N/C pushing toward positive x

For the -2C charge

E = (9 X 109)(-2 X 10-6)/(.025)2 = 2.88 X 107 N/C pulling toward positive x

The net is 2.88 X 107 N/C + 2.88 X 107 N/C =

5.76 X 107 N/C directed toward positive x

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