Question

7-11 please

7. Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2,000 N/C. If the voltage is halved and the distance between the plates is reduced to 1/4 the original distance, the magnitude of the new electric field is (A) 8000 N/C (B) 1,600 N/C (C) 2,400 N/C (D) 4,000 N/C (E) 10,000 N/C 8. A parallel-plate capacitor is charged by connection to a battery. If the battery remains connected and the separation between the plates is increased, what will happen to the charge on the capacitor and the voltage across it? (A) Both remain fixed. (B) Both increase. (C) Both decrease (D) The charge increases and the voltage decreases. (E) The charge remains fixed and the voltage increases.

Answer 1

7 same E- E=V option is correct d The change remains for new electric field and voltage VIL increases. dlu E': 2V 9 Emf - I (P + R2 + R3) d E'. 26 Emf = 05 ( 16+90+4) E' - 2X2000N/C Emf - 0.sx yo 4ooon E = 20 Volt 4000 NIC option E is correct option d is correct. 10) Vxy = E-IXY - 20-0.5xy C = E A - 202.0 = 18 volt due to increase ind IB Volt decreases. option is correct battery is disconnected so Pur I²R change will be fixed = 0.5²x4 v=g I watt V will increase due to 1 watt decrease in C. option D is correct. 8 EA E

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