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The field just outside a 4.00 cm radius metal ball is 3.25 102 N/C and points toward the ball. What charge resides on the ball?

My work:

r=4.00 cm --> 4.0x10^-2
E=3.25x10^2

Q=(3.25x10^2)[(4.0x10^-2)^2] divided by (9x10^9)
Q=(325)(.0016)/9x10^9
Q=5.7e-11

I submitted and I am wrong? Hmmm....

2007-08-29 09:53:36 · 7 answers · asked by Jessie L 2 in Science & Mathematics Physics

7 answers

The problem specified the direction of the electric field, which is toward the ball. The sign of the charge on the ball is therefore negative, so you could have lost points for getting the algebraic sign wrong. The magnitude is about right if the units of charge are coulombs, but you really should have specified the units too.

2007-08-29 10:07:55 · answer #1 · answered by devilsadvocate1728 6 · 2 0

Ok. Gauss' Law says that the field integrated over a closed surface is equal to the net charge enclosed in the surface. So

E* 4*pi*r^2 = q/ep where ep =8.85e-12 MKS units

So q = 4*pi*ep*r^2 *E= 4*pi*8.85e-12*(0.04)^*3.25x10^2
q = 5.76 x 10^-11 coul.

Probably were marked wrong because you left off units?

2007-08-29 17:02:34 · answer #2 · answered by nyphdinmd 7 · 0 1

It's the sign. The charge is NEGATIVE since the ball acts as a sink for the electric field. the magnitude is correct though

2007-08-29 18:07:36 · answer #3 · answered by marcus101 2 · 1 0

E = q/(4 pi e0 r^2), with q required. Rearranging, q = 4 pi e0 r^2 E, or 12.56 (8.85E-12) (.0016) 325. And I got the same answer you did -- as did the first responder.

2007-08-29 17:04:52 · answer #4 · answered by Anonymous · 0 1

Are you using the formula:
f=(k*q1*q2)/r^2 ; f is in newtons not n/c

Dimensional Analysis:
q=(n/c*cm)/(n*m^2/c^2)
=(n/c*cm)*( [c^2]/ [n*m^2] )
=c*((0.01)*m/m^2) ; n is gone, 1 c is gone
=c/m -> coulombs/meter

You have an extra meters unit in the denominator.

I need to brush up on my fields.

2007-08-29 17:11:35 · answer #5 · answered by active open programming 6 · 0 0

its correct answer but the sign of the charge is negative due to the direction of electric field

2007-08-30 04:25:26 · answer #6 · answered by MOHAMMAD A 1 · 0 0

You have to convert your centimeters to meters so that your units match up

2014-01-17 13:40:42 · answer #7 · answered by Craig 1 · 0 0

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