Question

6-11 please !!!

6. A positive charge of 6.0 x 10 coulomb is placed in an upward directed uniform electric field of 4.0 x 10^N/C. When the charge is moved 0.5 meter upward, the work done by the electric force on the charge is (A) 6 x 104J (B) 1.2 x 10°) (C)2 x 10^ J (D) 8 x 10^J (E) 12 x 10- J 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 NC (B) 1,600 N/C (C) 2,400 NC (D) 4,000 NC (E) 10,000 NC 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. 16 W X E I-0.5 A 200 Intemal Resistance Questions 9-11 relate to the following circuit diagram which shows a battery with an internal resistance of 4.0 ohms connected to a 16-ohm and a 20-ohm resistor in series. The current in the 20-ohm resistor is 0.5 amperes 9. What is the emf of the battery? (A)2 V (B)10V (C) 18 V (D) 12.0V (E) 20 V What is the potential difference across the terminals X and Y of the battery? (A) 20 V (B)2 V (C) 18 V (D) 12.0V (E) 22 V 11. What power is dissipated by the 4-ohm internal resistance of the battery? (A) 0.36 W (B) 1.2 W (C)3.2 W (D) IW (E) 0.5 W 10.

Answer 1

1) work done :

W = F*d = q*E*d

W = 6*4*10^-4*0.5

W = 12*10^-4 joules

2) E = V/d

So when voltage is halved and distance is 1/4 then, electric field will be twice of original value.

E = 4000 N/C

3) C = A $\varepsilon$ /d

If d is increased then, C is decreased and

Q = C*V

Charge will drcrease and voltage will remain constant as battery remains connected.

4) e = 0.5*(20+16+4)

e = 20 V

6) p = i^2r

p = 0.5^2*4

p = 1 Watt

** there is Y missing in the picture so can't find out the question 10