
Two charges repel each other with force F when they are at the distance a part.If charge is halved and distance is also halved?
Answer
219.3k+ views
Hint: The interaction between charged objects is a non-contact force that acts at some distance of separation of Charge i.e., without coming physically in contact .From Coulomb's law of electrostatic force we know,
${\text{F = }}\dfrac{{{{kq_1q_2}}}}{{{{\text{r}}^{\text{2}}}}}$
Complete step by step solution:
As given the distance between the charges is halved, with also charges, halved new distance becomes
$\eqalign{
& \Rightarrow {\text{F' = }}\dfrac{{{{k(q_1/2)(q_2/2)}}}}{{{{(r/2)}^2}}} \cr
& \Rightarrow {\text{F' = 4k}}\dfrac{{{{q_1q_2}}}}{{4{r^2}}} \cr
& \Rightarrow {\text{F' = }}\dfrac{{kq_1q_2}}{{{r^2}}} \cr
& \therefore {\text{F = F'}} \cr} $
Additional information:
Electric force also has a magnitude or power. Like most types of forces, there are a variety of factors that affect the magnitude of the electric force. Two types of charged balloons will repel each other and the force of their repulsive force can be changed by changing three variables. First, the amount of charge on one of the balloons will affect the force of the repulsive force. The more charged the balloon, the greater the repulsive force.
Second, the amount of charge on the second balloon will affect the force of the repulsive force. Gently rub two balloons with the animal's fur and they slightly retract. Rub both balloons vigorously so that both of them are charged more, and they move back a lot. Finally, the distance between the two balloons will have a significant and noticeable effect on the repulsive force. The electric force is strongest when the balloons are close together. The force increases as the separation distance decreases.
Note: The force between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between them. Therefore, if the distance between charges is halved (charges are kept constant), the force between the two charges is quadrupled.
${\text{F = }}\dfrac{{{{kq_1q_2}}}}{{{{\text{r}}^{\text{2}}}}}$
Complete step by step solution:
As given the distance between the charges is halved, with also charges, halved new distance becomes
$\eqalign{
& \Rightarrow {\text{F' = }}\dfrac{{{{k(q_1/2)(q_2/2)}}}}{{{{(r/2)}^2}}} \cr
& \Rightarrow {\text{F' = 4k}}\dfrac{{{{q_1q_2}}}}{{4{r^2}}} \cr
& \Rightarrow {\text{F' = }}\dfrac{{kq_1q_2}}{{{r^2}}} \cr
& \therefore {\text{F = F'}} \cr} $
Additional information:
Electric force also has a magnitude or power. Like most types of forces, there are a variety of factors that affect the magnitude of the electric force. Two types of charged balloons will repel each other and the force of their repulsive force can be changed by changing three variables. First, the amount of charge on one of the balloons will affect the force of the repulsive force. The more charged the balloon, the greater the repulsive force.
Second, the amount of charge on the second balloon will affect the force of the repulsive force. Gently rub two balloons with the animal's fur and they slightly retract. Rub both balloons vigorously so that both of them are charged more, and they move back a lot. Finally, the distance between the two balloons will have a significant and noticeable effect on the repulsive force. The electric force is strongest when the balloons are close together. The force increases as the separation distance decreases.
Note: The force between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between them. Therefore, if the distance between charges is halved (charges are kept constant), the force between the two charges is quadrupled.
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