Problem 3 Near the surface of the Earth there is an electric field of about 150

Question

Problem 3

Near the surface of the Earth there is an electric field of about 150 V/m which points downward. Two identical balls with mass m = 0.550 kg are dropped from a height of 2.10 m , but one of the balls is positively charged with q1 = 750 C , and the second is negatively charged withq2 = -750 C .

Part A

Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

Problem 4

A parallel-plate capacitor with plate area A = 8.0 m2and plate separation d = 3.0 mm is connected to a 35-V battery (refer to (Figure 1) ).

E0=1.17 x 10^4 V/m

C0= 2.36 x 10^-8 F

Q0= 8.26 x 10^-7 C

PE0= 1.45 x 10^-5 J

Part E

With the capacitor still connected to the battery, a slab of plastic with dielectric strength K = 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (refer to (Figure 2) ). What are the new value of electric field in the capacitor?

Part F

What are the new value of the capacitance?

Part G

What are the new value of thethe charge on the capacitor.

Part H

What are the new value the energy stored in the capacitor?

Explanation / Answer

I am allowed to answer only 1 question at a time

3)
Decrease in potential energy of positive charge = m*g*h + q*V
=0.550*9.8*2.1 + (750*10^-6)*150
=11.319 + 0.1125
=11.43 J
This is increase in kinetic energy of positive charge
KE1 = 11.43 J
0.5*m*V1^2 = 11.43 J
0.5*0.55*V1^2 = 11.43 J
V1= 6.45 m/s

Decrease in potential energy of Nehative charge = m*g*h + q*V
=0.550*9.8*2.1 + (-750*10^-6)*150
=11.319 - 0.1125
=11.21 J
This is increase in kinetic energy of negative charge
KE1 = 11.21 J
0.5*m*V2^2 = 11.21 J
0.5*0.55*V2^2 = 11.21 J
V2= 6.38 m/s

difference in speed = V1 - V2
= (6.45 - 6.38) m/s
= 0.07 m/s
Answer: 0.07 m/s

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