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:**

E = k|q’| / r^{2}

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

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

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

^{−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?

Enter your answer numerically in newtons per coulomb.

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

E = kq / r^{2}

Where k is the Coulomb constant (8.9875 * 10^{9})

So:

E = kq / r^{2}

E = (8.9875 * 10^{9}) * (1.62*10^{−6}) / (1.53)^{2}

E = 6220 N/C

Part D

Enter the letter of the vector that represents the direction of E->

_{P}.

G

Part E

^{-19}C.

Enter your answer numerically in newtons.

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

F = kq_{1}q_{2} / r^{2}

F = (8.9875 * 10^{9}) * (1.62*10^{−6}) * (1.60 * 10^{-19}) / (1.53)^{2}

9.95 * 10^{-16} N

Part F

Enter the letter of the vector that represents the direction of F->.

C

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