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JEE Corner Assertion and Reason 1. Capacitance =
q is constant for a given V
capacitor. 2. Reason correctly explains the assertion. 1 3. U = qV , W = qV 2 4. For discharging of capacitor q = q0e - t / t q dq = - 0 e -t / t t dt q = - 0 e -t / t RC Hence, more is the resistance, less will be the slope. 5. Charge on two capacitors will be same only if both the capacitors are initially uncharged.
6. As potential difference across both the capacitors is same, charge will not flow through the switch. 7. C and R2 are shorted. 8. Time constant for the circuit, t = RC 9. In series, charge remains same q2 1 and U= ÞU µ 2C C 10. In series charge remains same q q V1 = , V2 = \ C1 C2 On inserting dielectric slab between the plates of the capacitor, C2 increases and hence, V2 decreases. So more charge flows to C2.
Objective Questions (Level 2) ì 4Q ^ i ïï e0 A ï 2Q ^ 1. E = íi ï e0 A ï 4Q ^ ï e Ai î 0
for x < d
But, I01 = I02 Þ
for d < x < 2d Also, for 2d < x < 3 d
2. Let E0 = external electric field and E = electric field due to sheet E1 = E0 - E = 8 \ E2 = E0 + E = 12 s Þ E = 2 V/m Þ =2 2e0 s = 4e0 3. When the switch is just closed, capacitors behave like short circuit, no current pass through either 6 W or 5 W resistor. 4. For charging of capacitor I = I0 e - t / t
t t V t ln I = ln R RC ln I = log I0 -
V1 V2 = R1 R2
1 1 > R1C1 R2C2
Þ R2C2 > R1C1 As only two parameters can be different, C1 = C2 R2 > R1 and V2 > V1 5. Charge on capacitor at the given instant. q CE q= 0 = 2 2 Heat produced = Energy stored in capacitor q2 CE2 = = 2C 8 Heat liberated inside the battery, r = ´ Total heat produced r + 2r =
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