Definitely not possible. Capacitance is given by an expression
C = epsilon x A / d
Since charge is not present the capacitance cannot be increased or decreased by the charge placed
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.
If the distance between charges is increased while the charges remain the same, the force between the charges decreases. This relationship is described by Coulomb's law, which states that the force between charges is inversely proportional to the square of the distance between them.
By increasing the distance between them by sqrt(2).
It is increased by a factor of 2
The electric potential energy between two charged particles is directly proportional to the product of their charges. If one particle's charge is increased by a factor of 2, the potential energy between the two particles will increase by a factor of 2 as well.
If the charges increase, the force increases.If the charges increase, the force increases.If the charges increase, the force increases.If the charges increase, the force increases.
Increasing the number of resonance structures increases the delocalization of electrons, spreading the positive and negative charges over a larger area. This increased delocalization results in lower overall energy and increased stability of the molecule.
Increased Protein and increased carboyhydrates.
It is simple that if the selling price is increased more then of cost increase then profit will increase but if selling price increased less then cost increased then there will be less profit or selling price increased in same proportion to cost increased then there may be no increase in profit. Besides that there may be so many other reasons for that.
an increase in current, hence increased power and increased heat.
If 8 is increased to 12, the increase is what percentage of the original number?
in FV analogy, mass is analogous to inductance
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.
If the distance between charges is increased while the charges remain the same, the force between the charges decreases. This relationship is described by Coulomb's law, which states that the force between charges is inversely proportional to the square of the distance between them.
The past tense of increase is increased.
8.94% increase.
It is an increase of 79, which is an increase of about 121.54%.