Find electric force on a point charge

Please refer to the attached files.

*I only need solutions for these *

1. What is the magnitude of the force on a change of +2 x 10-7C that is 0.3m from a charge of -5 x 10-7C?
Answer: 10-2N attractive force.

Solution:
The electric force of a point charge as a result of the presence of another point charge is defined by the following equation:
F=(kq_1 q_2)/r^2
=(9.0×〖10〗^9 N m^2 C^(-2)×2×〖10〗^(-7) C×5×〖10〗^(-7) C)/〖(0.3m)〗^2
=〖10〗^(-2) N

Because the two point charges are of opposite charges, the force between them is attractive. Therefore, there is a 10-2 N attractive force on the +2×10-7C point charge.

2. Two electrons repel each other with a force of 10-8N. How far apart are they?
Answer: 1.5 x 10-10m

Solution:
The charge of an electron is -1.60 × 10-19 C. We can find the distance between the two electrons with this equation:
F=(kq_1 q_2)/r^2
〖10〗^(-8) N=(9.0×〖10〗^9 N m^2 C^(-2)×1.6×〖10〗^(-19) C×1.6×〖10〗^(-19) C)/r^2
r^2=(9.0×〖10〗^9 N m^2 C^(-2)×1.6×〖10〗^(-19) C×1.6×〖10〗^(-19) C)/(〖10〗^(-8) N)
r=1.5×〖10〗^(-10) m
Therefore, the distance between the two electrons is 1.5 × 10-10 m.

 
3. A test charge of +2 x 10-7C is located 5cm to the right of a charge of +1 x 10-6C and 10cm to the left of a charge of -1 x 10-6C. The three charges lie on a straight line. Find the force on the test charge.
Answer: 0.9 to the right

Let's find the forces one by one.
The force on the test charge +2×10-7 C due to the +1×10-6 C is repulsive and it is towards the right.

F=(kq_1 q_2)/r^2
=(9.0×〖10〗^9 N m^2 C^(-2)×2×〖10〗^(-7) C×1×〖10〗^(-6) C)/〖(0.05m)〗^2
=0.72N

The force on the test charge +2×10-7 C due to the -1×10-6 C is attractive and it is towards the right.

F=(kq_1 q_2)/r^2
=(9.0×〖10〗^9 N m^2 C^(-2)×2×〖10〗^(-7) C×1×〖10〗^(-6) C)/〖(0.1m)〗^2
=0.18N

Now, since both forces are towards the right, we can add them together.
The total force is = 0.72 N + 0.18 N
= 0.90 N to the right

 
4. Three +100-statC charges are arranged in a straight line, the second charge (B) is 20cm to the right of the first (A) and the third (C) is 50cm to the right of A. What is the force exerted by charges A and B on C?
Answer: 15.1 dyu

Now, this equation is stated in the CGS units.
F=(kq_1 q_2)/r^2
In CGS units, k = 1, q is in the unit of statC, r is in unit of cm.

The force on C due to A is
F=(kq_1 q_2)/r^2
=1(100)(100)/〖50〗^2 dynes
=4 dynes
This force is repulsive and is towards the right.

The force on C due to B is
F=(kq_1 q_2)/r^2
=1(100)(100)/〖30〗^2 dynes
=11.1 dynes
This force is also repulsive and is towards the right as well.

The total force is
F = 4 dynes + 11.1 dynes
= 15.1 dynes

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