A parallel-plate capacitor has plates of area 102.0 cm 2 and a separation of 1.3
ID: 2136830 • Letter: A
Question
A parallel-plate capacitor has plates of area 102.0 cm2 and a separation of 1.30 cm. A battery charges the plates to a potential difference of 172.0 V and is then disconnected. A dielectric slab of thickness 7.67 mm and dielectric constant 5.30 is then placed symmetrically between the plates. Find the capacitance before the slab is inserted.
What is the capacitance with the slab in place?
What is the free charge (charge on a plate) before the slab is inserted?
What is the free charge after the slab is inserted?
What is the magnitude of the electric field in the space between the plates and dielectric?
What is the magnitude of the electric field in the dielectric?
With the slab in place, what is the potential difference across the plates?
What is the magnitude of the external work involved in the process of inserting the slab?
Explanation / Answer
a. Co = A* Epsilon/d
= 0.0102* 8.85* 10^-12/ 0.013
= 6.94 * 10^-12 Farad
b. Before slab is inserted free charge = Co* V
= 6.94 * 10^-12 *172
= 1194 * 10^-12 C
c. After slab is inserted the charge ll remain the same..
d Electric Field = q/A* epsilon Not
= 1194*10^-12/.0102*8.85*10^-12
= 1.332* 10^4 V/m
e. New Electric field E'= 1.332*10^4/5.3
= 2.495*10^3 V/m
f. Potential diffenence = E(d-t) + E't
= 1.332* 10^4(0.013-0.00767) + 2.495*10^3(0.00767)
= 70.996+ 19.1367 = 90.1327 Volt
g. Work done = q^2/2(1/C - 1/Co)
=1/2(1194 * 10^-12)^2*(1/C - 1/Co)
we need to calculate C first
C = 8.85*106-12* 0.0102*5.3/{5.3(0.013-0.00767)}+(0.00767)
= 13.29 *10^-12 farad
Hence C= 13.29 *10^-12 farad
Co = 6.94 * 10^-12 Farad
subxtitute the values u ll get the answer...