Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`         A parallel-plate capacitor of capacitance 5 μF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?`
3 years ago

396 Points
```										Sol. C = 5 μf V = 6 V d = 2 mm = 2 × 10^–3 m.
(a) the charge on the +ve plate
q = CV = 5 μf × 6 V = 30 μc
(b) E = V/d = 6V/2 * 10^-3m = 3 * 10^3 V/M
(c) d = 2 * 10^-3 m
T = 1 * 10^-3 m
k = 5 or C = ε base 0A/d ⇒ 5 * 10^-6 = 8.85 * A * 10^-12/2 * 10^-3 * 10^-9 ⇒ A = 10^4/8.85
When the dielectric placed on it
C base 1 = ε base 0A/d-t+t/k = 8.85 * 10^-12 * 10^4/8.85/10^-3 + 10^-3/5 = 10^-12 * 10^4 * 5/6 * 10^-3 = 5/6 * 10^-5 = 0.00000833 = 8.33 μF.
(d) C = 5 × 10^–6 f. V = 6 V
∴ Q = CV = 3 × 10^–5 f = 30 μf
C’ = 8.3 × 10^–6 f
V = 6 V
∴ Q’ = C’V = 8.3 × 10^–6 × 6 ≈ 50 μF
∴ charge flown = Q’ – Q = 20 μF

```
3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Electromagnetic Induction

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Electricity and Magnetism
• OFFERED PRICE: Rs. 1,696
• View Details