The field is
<em><u>E</u></em> = 1 / (4 pi ε₀) Q / <em><u>R</u></em>² directed radially outward from
the center of the shell.
R is the radius of the spherical shell.
Notice that the field is exactly the same as the field due to a point-charge
with magnitude 'Q' that's located at the center of the sphere.
Answer:
9 - 10N to the left
10 - There is no change on the object
Explanation:
Can I have brainliest answer pls?
The answer would be E7. Galaxies categorized as E0 look to
be nearly perfect, while those registered as E7 seem much extended than they
are widespread. It is worth noting, though, that a galaxy's look is connected
to how it lies on the sky when viewed from Earth. An E7 galaxy is very long and
thin or the flattest of them all.
Answer:
Q1 = 7.25*10^(-16) C
Explanation:
We are given;
electric field strength = (1 x 10^5 N/C
drag force (F) = 7.25 x 10^(-11) N
The question says it's moving with constant velocity. This means that he particle is in equilibrium and not accelerating.
Columbs law force of attraction or repulsion between two charges is given as;
F=(KQ1Q2)/r²
Now, electric field strength is given as the formula;(K*Q2)/r², thus plugging the relevant values gives us;
7.25 x 10^(-11) N= (1 x 10^(5) N/C)Q1 Q1 = 7.25 x 10^(-11) /(1 x 10^(5))
Q1 = 7.25*10^(-16) C
