1) a) A charge of +5.00 nC is placed at the origin, and another charge of -3.00

Question

1) a) A charge of +5.00 nC is placed at the origin, and another charge of -3.00 nC is

placed 2.5 cm west of origin, what is the electric potential at 3 cm east from origin?

b) Repeat a) at distance 5 cm north of origin.

c) Find the work done to bring another charge of -5.00 nC from infinity to the point at 3 cm east from origin.

with the same three capacitors in parallel. How is this different than the result in series?

2) Three resistors of 25.00 ?, 17.00 ? and 13.00 ? are connected in series to a source with a potential difference of 3.00 V.

a) What is the equivalent resistance?

b) What is the the current in each resistor, and what is the potential difference across each resistor?

c) What is the power in each capacitor?

with the the same resistors in parallel. How is this different than the result in series?

Explanation / Answer

Number 1

Part A)

The formula for electric potential is V = kq/r

So we need to apply that for both charges

V = (9 X 109)(5 X 10-9)/(.03) + (9 X 109)(-3 X 10-9)/(.055)

V = 1010 V

Part B)

The distance from the negative charge to the location in question is found from the Pythagorean Theorem

d2 = (52) + (2.5)2

d = 5.59 cm

V = (9 X 109)(5 X 10-9)/(.05) + (9 X 109)(-3 X 10-9)/(.059)

V = 442.4 V

Part C)

W = q(delta V)

W = (-5 X 10-9)(1010)

W = -5.05 X 10-6 J

Number 2)

Part A)

Resistance in series adds algebraically

R = 25 + 17 + 13

R = 55 Ohms

Part B)

Current will be the same since current is constant in series

V = IR

3 = I(55)

I = .0545 A (54.5 mA) in all resistors

For the Voltage apply V = IR separately

V = (.0545)(25) = 1.36 V across the 25 Ohm

V = (.0545)(17) = .927 V across the 17 Ohm

V = (.0545)(13) = .709 V across the 13 Ohm

Part C)

Asks about capacitors? Makes no sense since we are talking abour resistors.

But Power in the Resistors is...

P = I2R

For the 25 Ohm P = (.0545)2(25) = .0744 Watts

For the 17 Ohm P = (.0545)2(17) = .0506 Watts

For the 13 Ohm P = (.0545)2(13) = .0387 Watts

FOR PARALLEL DIFFERENCE

In parallel voltage is constant and current splits

To find the resistance, it adds inversely...

1/R = 1/25 + 1/17 + 1/13

R = 5.69 Ohms

The voltage across them all is 3V

For current...

V = IR

For the 25 Ohm, I = V/R = 3/25 = .12 Amps

For the 17 Ohm, I = V/R = 3/17 = .176 Amps

For the 13 Ohm, I = V/R = 3/13 = .231 Amps

For the power

The 25 Ohm... P = I2R = (.12)2(25) = .36 Watts

The 17 Ohm... P = I2R = (.176)2(17) = .527 Watts

The 13 Ohm... P = I2R = (.231)2(13) = .694 Watts

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