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What am I doing wrong physics lovers?
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....
7 Answers
- 1 decade agoFavorite Answer
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.
- nyphdinmdLv 71 decade ago
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?
- 1 decade ago
It's the sign. The charge is NEGATIVE since the ball acts as a sink for the electric field. the magnitude is correct though
- Anonymous1 decade ago
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.
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- 1 decade ago
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.
- 1 decade ago
its correct answer but the sign of the charge is negative due to the direction of electric field