Question

4. Think back to Coulomb's Law. Two coins with identical charges are placed on a lab table 1.35 m apart.
A. If each coin experiences a force of 2.0 N because of the presence of the other coin, calculate the charge on each coin.
B. Would the force be classified as a force of attraction or repulsion? Explain your answer.

Answer 1

A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:
 
 F=K(q1xq2)/r²
 
 F: The magnitud of the force between the charges. (F=2.0 N).
 K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
 q1 and q2: Electrical charges.
 r: The distance between the charges (r=1.35 m).
 
 We have the values of F, K and r, so we can calculate q1xq2, because both coins have  identical charges:
 
 q1xq2=(r²xF)/K
 q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
 q1xq2=3x10^-10 C
 q1=q2=(3x10^-10 C)/2
 
 Then, the charge of each coin, is:
 
 q1=1.5x10^-10 C 
 
 q2=1.5x10^-10 C

B) Would the force be classified as a force of attraction or repulsion?
 
 It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.

Answer 2

Answer:

A.

$q = 2 \times 10^{-5} C$

B. may be attractive or repulsive

Explanation:

Let the charge on each coin is q. The separation between the coins is d.

According to the Coulomb's law, the force between the two coins is given by

$F = K \frac{q \times q}{d^{2}}$

A.

According to the question, F = 2 N, d = 1.35 m

So,

$2 = 9 \times 10^{9} \frac{q \times q}{1.35^{2}}$

By solving,

$q = 2 \times 10^{-5} C$

B.

The nature of force can be determined by the nature of charges. If the charge on both the coins is same in nature, then the force is repulsive and if the charge on both the coins be of opposite nature, then the force is attractive in nature.