## Mastering Physics Solutions: Electric Fields and Forces

Electric Field Conceptual Question

Part A =
the distance between the positive charge and the electron
the charge on the electron
the charge of the positive charge
Part B =
the distance between the positive charge and the electron
the charge of the positive charge
Part C = 6220 N/C
Part D = G
Part E = 9.95 * 10-16 N
Part F = C

Solutions Below:

An electric field can be created by a single charge or a distribution of charges. The electric field a distance r from a point charge q’ has magnitude

E = k|q’| / r2

The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge q is placed in an electric field created by q’, q will not significantly affect the electric field if it is small compared to q’.

Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).

Part A

There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?
Check all that apply.

• the distance between the positive charge and the electron
• the charge on the electron
• the mass of the electron
• the charge of the positive charge
• the mass of the positive charge
• the radius of the positive charge
• the radius of the electron

the distance between the positive charge and the electron, the charge on the electron, the charge of the positive charge

Part B

For the same situation as in Part A, on which of the following quantities does the electric field at the electron’s position depend?
Check all that apply.

• the distance between the positive charge and the electron
• the charge on the electron
• the mass of the electron
• the charge of the positive charge
• the mass of the positive charge
• the radius of the positive charge
• the radius of the electron

the distance between the positive charge and the electron, the charge of the positive charge

Part C

If the total positive charge is Q = 1.62*10−6 C, what is the magnitude of the electric field caused by this charge at point P, a distance d = 1.53 m from the charge?

The magnitude of an electric field is given by the formula:

E = kq / r2

Where k is the Coulomb constant (8.9875 * 109)

So:

E = kq / r2
E = (8.9875 * 109) * (1.62*10−6) / (1.53)2

E = 6220 N/C

Part D

What is the direction of the electric field at point P?
Enter the letter of the vector that represents the direction of E->P.

G

Part E

Now find the magnitude of the force on an electron placed at point P. Recall that the charge on an electron has magnitude e = 1.60 * 10-19 C.

The force is just the electric effect of two charges. We can find it using a similar formula as Part C:

F = kq1q2 / r2
F = (8.9875 * 109) * (1.62*10−6) * (1.60 * 10-19) / (1.53)2

9.95 * 10-16 N

Part F

What is the direction of the force on an electron placed at point P?
Enter the letter of the vector that represents the direction of F->.

C

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### 10 Responses to Mastering Physics Solutions: Electric Fields and Forces

1. Kristen says:

This made so much sense; thank you so much!

• Mastering Physics Solutions says:

Glad to help, you’re very welcome!

2. Ashley says:

• Mastering Physics Solutions says:

You’re very welcome!

3. Branden says:

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• Mastering Physics Solutions says:

Thank you for the kind remarks! You’re very welcome!

4. ben says: