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If the distance between the charges doubles, the force between them decreases by a factor of 4. This relationship is described by Coulomb's law, which states that the force is inversely proportional to the square of the distance.

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What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is double?

If the magnitude of both charges is doubled and the distance between them is also doubled, the force between them will remain the same. This is because the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Doubling both charges and distance cancels each other out in terms of force.


If you double the distance between two objects the electric force is .?

If you double the distance between two objects, the electric force between them decreases by a factor of four. This is because electric force is inversely proportional to the square of the distance between the charges.


What happens to the force between two charges if the magnitude of both charges is doubled and the distance between is doubled?

Force of attraction between charges is directly proportional to the charge. So as we quadrule each charge then force will become 4x4 ie 16 times increased Force is also inversely related to the square of the distance. So as we double the distance then the force is decreased by 22 ie 4 times Hence the net change will be 16/4 ie 4 times increase in the force of attraction.


If you double the distance between two charged objects by what factor is the electric force affected?

If you double the distance between two charged objects, the electric force between them decreases by a factor of four. This is because the electric force is inversely proportional to the square of the distance between the charges according to Coulomb's Law.


Two charges are separated by a distance d If one the charges is doubled while the other is tripled what happens to the force between the charges?

We have to assume that the distance between the charges remains constant, and the answer doesn't depend on the distance. The force between the charges is proportional to the product of the charges. Initial force = constant x (Q1) x (Q2) New force = constant x (2Q1) x (3Q2) = 6 x (Q1) x (Q2) = 6 times the initial force. The direction of the force doesn't change. It's attractive if the charges are of opposite sign, repulsive if they're of like sign.

Related Questions

Determine the amount by which the electric force between two charges is increased when the distance between the charges is halved?

If the charge on the object is double than the force between them is double


What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is double?

If the magnitude of both charges is doubled and the distance between them is also doubled, the force between them will remain the same. This is because the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Doubling both charges and distance cancels each other out in terms of force.


If you double the distance between two objects the electric force is .?

If you double the distance between two objects, the electric force between them decreases by a factor of four. This is because electric force is inversely proportional to the square of the distance between the charges.


What happens to the force between two charges if the magnitude of both charges is doubled and the distance between is doubled?

Force of attraction between charges is directly proportional to the charge. So as we quadrule each charge then force will become 4x4 ie 16 times increased Force is also inversely related to the square of the distance. So as we double the distance then the force is decreased by 22 ie 4 times Hence the net change will be 16/4 ie 4 times increase in the force of attraction.


What happenes to the force between two charges if the magnitude of both is doubled and the distance between them is doubled?

Force of attraction between charges is directly proportional to the charge. So as we quadrule each charge then force will become 4x4 ie 16 times increased Force is also inversely related to the square of the distance. So as we double the distance then the force is decreased by 22 ie 4 times Hence the net change will be 16/4 ie 4 times increase in the force of attraction.


If you double the distance between two charged objects by what factor is the electric force affected?

If you double the distance between two charged objects, the electric force between them decreases by a factor of four. This is because the electric force is inversely proportional to the square of the distance between the charges according to Coulomb's Law.


Two charges are separated by a distance d If one the charges is doubled while the other is tripled what happens to the force between the charges?

We have to assume that the distance between the charges remains constant, and the answer doesn't depend on the distance. The force between the charges is proportional to the product of the charges. Initial force = constant x (Q1) x (Q2) New force = constant x (2Q1) x (3Q2) = 6 x (Q1) x (Q2) = 6 times the initial force. The direction of the force doesn't change. It's attractive if the charges are of opposite sign, repulsive if they're of like sign.


What happens to the electric force if you double the distance between two objects?

The electric force between two objects decreases to one-fourth of the original force if the distance between them is doubled. This is because the electric force is inversely proportional to the square of the distance between the charges.


What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is cut in half?

The force between the two charges increases 16 times.Coulomb's law equation states:│F│ = ke │q1q2│/r2Where (F) is the force acting simultaneously on both point charges (q1) and (q2).r is the separation distance and ke is a proportionality constant called the Coulombconstant.Using above equation, if we double both charges and reduce the distance in half.│F2│ = ke │(2q12q2)│/(r/2)2 = 16 ke │q1q2│/r2 = 16 │F│.We see that the force turns out 16 times stronger


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