Question

Draw the electric field around a +2.0 × 10–5 C point source charge. Draw the field lines and indicate their direction. Draw two different equipotential lines.

Answer 1

B. Calculate the force between a +5 µC test point charge and this source charge at a distance of 2.00 cm. (µC = 1.0 × 10–6 C)

Answer 2

C. If the test charge were moved closer to the source charge, would the change in its potential energy be positive, negative or zero? Explain.

Answer 3

@Vandreigan

Answer 4

Electric field lines radiate outward from a point source.

Any point that is the same distance from the point source will have the same field, and the same potential (equipotential = same field).

Answer 5

Can you answer B and C for me?

Answer 6

For B:
\[F = k \frac{q_1q_2}{r^2}\]
with:
\[k = \frac{1}{4 \pi \epsilon_o}\]

Answer 7

Since both charges are positive, they repel one another, and the closer they get to one another, the harder they push against one another. If you let the test charge go, it would go flying away.

Since energy isn't created or destroyed, where did the kinetic energy (energy of the motion of the charge) come from? This energy had to already be there in the form of potential energy.

Answer 8

So whats the answers? I understand that positive and positive push away but I dont know how to answer B and calculate it

Answer 9

you plug what you have into the formula I gave you.

q1 and q2 are the charges

Answer 10

I don't understand where to plug it I know the F=K q1q2/r2 thing just not where to put it

Answer 11

k is just a number. q1 and q2 are the values of the two charges. r is the distance they are separated by

Answer 12

I don't get it would it be k 5 1 52/2^2

Answer 13

hello @Vandreigan

Answer 14

We're given:
q1 = 2x10^(-5)C
q2 = 5x10^(-6)C
r = 0.02 m

Plug everything in and find F

Answer 15

ugh

Answer 16

5*10^-9?

Answer 17

is that correct @Vandreigan

Answer 18

Not quite. What value are you using for k?

Answer 19

Can you please just show me how to do this? I didn't even put k in the equation you said thr q1 and q2 and r2 nothing about k can you show me so i know how

Answer 20

I already showed how to do it:
\[F = k \frac{q_1q_2}{r^2} = k \frac{(2 \times 10^{-5})(5 \times 10^{-6})}{(0.02)^2}\]
Plug in the value for k, and get your answer

Answer 21

I'm so confused please explain better

Answer 22

We have an equation for force from an electric field. All we have to do is plug things into it, while being careful of units. I've gone further than I normally like to do.

The value of k is something you should have access to. Find it, plug it in. That's your last step.

Answer 23

Calculate the force between a +5 µC test point charge and this source charge at a distance of 2.00 cm. (µC = 1.0 × 10–6 C) that is all i have!

Answer 24

It'll be in your book. That, or the value that you need to solve for it:
\[k = \frac{1}{4 \pi \epsilon_o}\]