A uniform electric field of 5 kN/Cis in the x direction. A positive point charge

ID: 1267049 • Letter: A

Question

A uniform electric field of 5 kN/Cis in the x direction. A positive point charge of 3 ?C is released from the rest at the origin.

a.) What is the potential difference V (18 m) ? V (0)?

Answer in units of kV

b.) What is the change in the potential energy of the charge from x = 0 to x = 18 m?

Answer in units of mJ

c.) What is the kinetic energy of the charge when it is at x = 18 m ?

Answer in units of mJ

d.) Find the potential if it is chosen to be zero at x = 0.

1. V (x) = ?(5 kV/m)x + 5 kV/m

2. V (x) = ?(5 kV/m)x ? 10 kV/m

3. V (x) = ?(5 kV/m)x + 10 kV/m

4. V (x) = ?(5 kV/m)x ? 5 kV/m

5. V (x) = ?(10 kV/m)x

6. V (x) = ?(5 kV/m)x

e.) Find the potential if it is 6 kV at x = 0.

1. V (x) = ?(5 kV/m)x - 6 kV

2. V (x) = ?(5 kV/m)x + 5 kV/m

3. V (x) = ?(6 kV)x

4. V (x) = ?(5 kV/m)x +6 kV

5. V (x) = ?(5 kV/m)x

6. V (x) = ?(5 kV/m) - 5 kV/m

f.) Find the potential if it is zero at x = 1 m.

1. V (x) = ?(5 kV/m)x ? 10 kV

2. V (x) = ?(10 kV/m)x

3. V (x) = ?(5 kV/m)x
4. V (x) = ?(5 kV/m)x ? 5 kV

5. V (x) = ?(5 kV/m)x + 10 kV

6. V (x) = ?(5 kV/m)x + 5 kV

(a)V(18)-V(0) = -E*d = -5*18 = -90 kV

(b) P.E = -E*d*q = -90,000 * 3*10^-6

= -270 mJ

(d) 6. V (x) = ?(5 kV/m)x

(e) 4. V (x) = ?(5 kV/m)x +6 kV

(f) 6. V (x) = ?(5 kV/m)x + 5 kV

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