# Distance, electric field and potential

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

#### David513

##### Full Member
5+ Year Member

Members don't see this ad.
Hey all,

Conceptually I'm missing something about how distance is related to electric field and potential. Electric field is force per unit charge and potential is force per unit charge applied over a distance. Right? Well, the force changes with distance by getting larger when objects are brought closer together (F=kqq/r^2). But potential gets smaller when the distance becomes smaller/the objects are brought closer together ... which doesn't make any sense because the force increases to a really significant extent when the objects are closer together. Shouldn't potential increase when the objects are closer? I'm so confused.
Field=E=F/q
Potential=V=F/q * d

#### victorias

##### Full Member
Force = kq1q2/r^2 The farther away the two charged objects are, the less the force between them
Field = kq/r^2 is the force exerted on a positive test charge at some point in space, the farther away the test charge is from the charged object, the lower the electric field
Potential = kq/r Potential is voltage that an object exerts at some point in space, the farther away from the charged object, the lower the potential will be

#### aldol16

##### Full Member
5+ Year Member
But potential gets smaller when the distance becomes smaller/the objects are brought closer together ... which doesn't make any sense because the force increases to a really significant extent when the objects are closer together. Shouldn't potential increase when the objects are closer? I'm so confused.

This is where you are confused. V = k*Q/r. Therefore, as r decreases, V must increase.

#### David513

##### Full Member
5+ Year Member
Thanks for the answers! Still a teensy confused but I think talking it out helped me.