It's B inertia if I remember correctly you can also rule out A,S and D so...
Answer:
A) 0 for r < R B) Q/4πε₀r² for r > R
Explanation:
Here is the complete question
Part A Find the electric field inside a hollow plastic ball of radius R that has charge Q uniformly distributed on its outer surface. Give your answer as a multiple of Q/ε0.
Part B Find the electric field outside this ball. Give your answer as a multiple of Q/ε0. Express your answer in terms of some or all of the variables R, r and the constant π.
Solution
Using Gauss' law ∫E.dA = q/ε₀. Where E is the electric field, dA is the area vector and q is the charge enclosed.
A For r < R The direction of the electric field is directed radially inward and r is outward and the angle between them is 180°. So E.dA = EdAcos180 = -EdA
∫-EdA = q/ε₀
-E∫dA = q/ε₀
-E4πr² = q/ε₀ (∫dA = 4πr² since it is a sphere)
E = -q/4πr²ε₀
But for r < R q = 0. So,
E = -q/4πr²ε₀ = -0/4πr²ε₀ = 0
B For r > R The direction of the electric field is directed radially outward and r is outward and the angle between them is 0°. So E.dA = EdAcos0 = EdA
∫EdA = q/ε₀ where Q is the charge on the hollow plastic ball
E∫dA = q/ε₀
E4πr² = q/ε₀ (∫dA = 4πr² since it is a sphere)
E = q/4πr²ε₀
But for r > R q = Q. So,
E = Q/4πr²ε₀ = Q/4πε₀r²
Answer:
The magnitude of the current in the circuit is 0.01159 A.
Explanation:
Given that,
Capacitor
Inductance
Voltage V = 2 V
We need to calculate the energy when the capacitor is charged
Using formula of energy
Put the value into the formula
When the capacitor is discharged
Using formula of energy
rms value of current,
Hence, The magnitude of the current in the circuit is 0.01159 A.
The electric force between two charges is:
F = (9 x 10⁹) Q₁ Q₂ / D²
F is the force, in Newtons
Q₁ and Q₂ are the two charges, in Coulombs
D is the distance between them, in meters
I'm going to assume that the first little 'c' in the question stands for "Coulombs", and the second little 'c' stands for "centimeters". And now, I'll proceed to answer the question that I've just invented.
For these two electrons:
F = (9 x 10⁹) (1.6 x 10⁻¹⁹) (1.6 x 10⁻¹⁹) / (5.4 x 10⁻⁷)²
F = (9 x 1.6 x 1.6 x 10⁻²⁹) / (5.4² x 10⁻¹⁴)
F = (23.04 / 29.16) x 10⁻¹⁵
<em>F = 7.9 x 10⁻¹⁶ of a Newton</em>
<em></em>
EACH electron feels that same force, pushing it away from the other electron.
Answer:
a. 1.0 eV
Explanation:
Given that
Voltage difference ,ΔV = 1 V
From work power energy
Work =Change in the kinetic energy
We know that work on the charge W= q ΔV
For electron ,e= 1.6 x 10⁻¹⁹ C
q=e= 1 x 1.6 x 10⁻¹⁹ C
Change in the kinetic energy
ΔKE= q ΔV
Now by putting the values
ΔKE= 1 x 1.6 x 10⁻¹⁹ x 1 C.V
We can also say that
ΔKE= 1 e.V
Therefore the answer will be a.
a). 1.0 eV