Answer:
Capacitance of the second capacitor = 2C
Explanation:

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.
Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.
We have

Similarly for capacitor 2

Capacitance of the second capacitor = 2C
Answer:


Explanation:
k = Coulomb constant = 
Q = Charge
r = Distance = 8 cm
R = Radius = 4 cm
Electric field is given by

Volume charge density is given by

The volume charge density for the sphere is 

The magnitude of the electric field is 
Answer:
0.0002 C.
Explanation:
Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)
Mathematically, charge can be expressed as
Q = CV ................................. Equation 1
Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.
Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.
Substitute into equation 1
Q = 2×10⁻⁶× 100
Q = 2×10⁻⁴ C
Q = 0.0002 C.
The amount of charge accumulated = 0.0002 C
The net force on the sledge is 31.64N.
Frictional force = µkR
= 0.269 x 42.2 x 9.81 = 111.36
net force = 143N - 111.36N
= 31.64N
refer brainly.com/question/24557767
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