Question

Fixed ChargesThe drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = + 8.42 μC; the other two charges have identical magnitudes, but opposite signs: q2 = - 5.76 μC and q3 = + 5.76 μC. The distance between q1 and the other two charges is 2.05 m. Determine the net force exerted on q1 by the other two charges.

Accepted Answer

This is related to couloms law.
The following figure will show the situation.

The open dot represent the q1 charge. The red dot represent the q2 and the black dot represent q3.
 
q1 and q2 are opposite charges. Then they will have an attraction. The atraction force is at   direction rarr in the q1 charge.
 
q1 and q3 are both positive charges.Then they will repel each other. Due to this q1 will have a repel force in rarr direction.
 
The electrostatic force(F) between two charges(q1 and q2) at a distance(r) is given by;
F = (K*q1*q2)/r^2 
k = 8.987xx10^9Nm^2C^(-2) 
 
For our question;
q1 = +8.42 muC = +8.42xx10^(-6) C 
q2 = -5.76 muC = -5.76xx10^(-6) C 
q3 = +5.76 muC = +5.76xx10^(-6) C 
 
Let;
F1 = Attraction force between q1 and q2
F2 = Repel between q1 and q3
 
rarrF1 = 8.42xx10^(-6)*5.76xx10^(-6)*8.987xx10^9/(2.05)^2 
rarrF1 = 0.1037 N 
 
rarrF2 = 8.42xx10^(-6)*5.76xx10^(-6)*8.987xx10^9/(2.05)^2
rarrF1 = 0.1037 N
 
Net force extend on q1 = rarrF1 + rarrF2 
                                  = 0.1037*2 
                                  = 0.2074 N 
 
Net force extend on q1 is 0.2074 N.

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