- #1
Punchlinegirl
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A charge of 2pC is uniformly distributed throughout the volume between concentric spherical surfaces having radii of 1.3 cm and 3.3 cm. What is the magnitude of the electric field 1.8 cm from the center of these surfaces? Answer in units of N/C.
I used the equation [tex] \Phi = E*4\pi r^2 [/tex] and then the equation [tex] \Phi = q_e_n_c /E_0 [/tex] and set them equal to each other. I used the formula q_e_n_c = charge/volume * volume to get [tex] Q/(4/3)\pi R^3 * (4/3)\pi r^3. [/tex] Simplifying gave me [tex] Qr^3/E_0 R^3 [/tex]. Plugging this in for q_e_n_c and solving for the E field gave me ,
[tex] E= Qr/ E_0R^3*4\pi [/tex]
Q= 2 x 10^-12 since the charge is uniform
r =.018 m
E_0 = 8.85 x 10^-12
R= .033
So 2 x 10^-12(.018)/8.85 x 10^-12 *(.033)^3 * 4pi = 9.02 N/C... which isn't right.
Can someone please help me? I have no idea what I'm doing.
I used the equation [tex] \Phi = E*4\pi r^2 [/tex] and then the equation [tex] \Phi = q_e_n_c /E_0 [/tex] and set them equal to each other. I used the formula q_e_n_c = charge/volume * volume to get [tex] Q/(4/3)\pi R^3 * (4/3)\pi r^3. [/tex] Simplifying gave me [tex] Qr^3/E_0 R^3 [/tex]. Plugging this in for q_e_n_c and solving for the E field gave me ,
[tex] E= Qr/ E_0R^3*4\pi [/tex]
Q= 2 x 10^-12 since the charge is uniform
r =.018 m
E_0 = 8.85 x 10^-12
R= .033
So 2 x 10^-12(.018)/8.85 x 10^-12 *(.033)^3 * 4pi = 9.02 N/C... which isn't right.
Can someone please help me? I have no idea what I'm doing.