Answer:
see the answers below
Explanation:
we can use a Gaussian's surface to calculate E:
[tex]\int E \cdot dS=\frac{Q}{\epsilon_0}\\\\E\int dS=E4\pi r^2=\frac{Q}{\epsilon_0}\\\\E=\frac{Q}{4\pi r^2 \epsilon_0}[/tex]
Where Q is the net charge inside the Gaussian' s surface
Hence, we have
a)
for r:
[tex]E=\frac{Q}{4\pi \epsilon_0 r^2}[/tex]
b)
for R1, the Gaussian's surface has a charge of -Q (the charge at the center):
[tex]E=\frac{-Q}{4\pi \epsilon_0 r^2}[/tex]
c)
for R2, the net charge inside the Gaussian's surface is -Q+4Q=+3Q, hence:
[tex]E=\frac{3Q}{4\pi \epsilon_0 r^2}[/tex]
HOPE THIS HELPS!!
The Electric field according to the scenario will be:
(a) [tex]E = \frac{Q}{4 \pi \epsilon_0 r^2}[/tex]
(b) [tex]E = -\frac{Q}{4 \pi \epsilon_0 r^2}[/tex]
(c) [tex]E = \frac{3Q}{4 \pi \epsilon_0 r^2}[/tex]
By using the Gaussian's surface,
→ [tex]\int E.dS = \frac{Q}{\epsilon_0}[/tex]
→ [tex]E \int dS = E4 \pi r^2[/tex]
[tex]= \frac{Q}{\epsilon_0}[/tex]
then,
→ [tex]E = \frac{Q}{4 \pi r^2 \epsilon_0}[/tex]
here,
(a)
The electric field for "r" will be:
→ [tex]E = \frac{Q}{4 \pi \epsilon_0 r^2}[/tex]
(b)
The electric field for "R1" will be:
→ [tex]E = -\frac{Q}{4 \pi \epsilon_0 r^2}[/tex]
(c)
The electric field for "R2" will be:
→ [tex]E = \frac{3Q}{4 \pi \epsilon_0 r^2}[/tex]
Thus the responses above is correct.
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