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....
devilsadvocate1728
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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.
nyphdinmd
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?
marcus101
It's the sign. The charge is NEGATIVE since the ball acts as a sink for the electric field. the magnitude is correct though
Anonymous
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.
active open programming
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.