Answer:
a) 4
b)
c) 6 volts
Explanation:
Given:
Initial potential difference, V₁ = 12 V
Potential difference after inserting the dielectric, V₂ = 3 V
Now,
a) The charge is given as, q = CV
where, C is the capacitance
thus,
without dielectric, q = C × 12 = 12C .........(1)
now, let the dielectric constant be 'k'
therefore,
q = kCV₂
or
12C = kC × 3 (q from 1)
or
k = 4
b) Now, the energy after inserting a dielectric with dielectric constant as k become the times the energy without the dielectric
thus,
Final energy = × Initial energy
or
Final energy = 0.25 × Initial energy
or
Final energy = 25% of the energy without dielectric
therefore,
hence, the energy loss fraction is
c) Now,
removing half way out the dielectric
thus,
k' = k/2
or
k' = 4/2 = 2
and no charge escape
thus,
12C =
or
V' =
or
V' = 6 volts
therefore,
the voltmeter will read 6 volts