Question

hey guys please help me out with this question

The electric field everywhere on the surface of a thin
spherical shell of radius 0.750 m is measured to be equal
to 890 N/C and points radially toward the center of the
sphere. (a) What is the net charge within the sphere’s
surface? (b) What can you conclude about the nature

Answer 1

@Luigi0210 @Nnesha

Answer 2

@sissyedgar @YanaSidlinskiy

Answer 3

idk!

Answer 4

NO prob ! ;) i need some help please someone help me out

Answer 5

@M4thM1nd

Answer 6

Ok, first of all, let's use the Gauss law:
\[EA = Q/\epsilon_0\], but remember that it is a spherical shell, not a sphere, that means that the inner portion is empty.

Answer 7

First let's use Gauss law for \(r \ge R\), where \(R\) is the spherical shell radius

Answer 8

\(E*(4\pi r^2)=Q/\epsilon_0\), so...
\[E = \frac{Q}{4\pi \epsilon_0 r^2}\]
For \(r \ge R\)

Answer 9

Now, for the inner part, we have no charges enclosed by a gaussian surface, so..
\(EA = 0 \Rightarrow E = 0\), for \(r < R\)

Answer 10

As the electric field points towards teh center, the charge on the surface is Negative,
using the fist case \(r \ge R\), we get:
\[Q = 4 \pi \epsilon_0 R^2E = 4 \pi \epsilon_0* 0.750^2*(-890) =...\]

Answer 11

remember that \(\epsilon_0 = 8.85\times 10^{-12} C^2/{N\cdot m^2}\)

Answer 12

Got it @rock_mit182 ?

Answer 13

@M4thM1nd what if there is a charge, in the center of an sphere, with an electric field pointing inwards, How would I calculate the net charge (q NET) ?

Answer 14

Can you be more specific? Is it a pontual charge in a spherical shell?

Answer 15

if so, again we use Gauss Law, \[EA = Q_{env}/\epsilon_0\]
\[E*(4\pi r^2) = Q_{env}/\epsilon_0\]
\[E = \frac{Q_{env}}{4\pi \epsilon_0 r^2}\], for \(r> 0\)

Answer 16

thanks for helping

Answer 17

No problem ;)