The relation between Electric Field and Electric Potential

[ItsAnshumaan]

Homework Statement

The electric field and the electric potential at a point are E and V respectively.
(a) If E=0, V must be 0
(b) If V=0, E must be 0
(c) If E≠0, V cannot be 0
(d) If V≠0, E cannot be 0

Homework Equations

[/B]
E = V/d

The Attempt at a Solution

[/B]
I basically substituted the value of E and V as 0 in respective cases, but ended up getting (a) and (b) as true. I know this is a very fundamental question, but I just can't figure it out.

 

[cnh1995]

In your question, V is the absolute potential at the given point while electric field E=potential difference/d.

 

[SammyS]

Homework Statement

The electric field and the electric potential at a point are E and V respectively.
(a) If E=0, V must be 0
(b) If V=0, E must be 0
(c) If E≠0, V cannot be 0
(d) If V≠0, E cannot be 0

Homework Equations

[/B]
E = V/d

The Attempt at a Solution

[/B]
I basically substituted the value of E and V as 0 in respective cases, but ended up getting (a) and (b) as true. I know this is a very fundamental question, but I just can't figure it out.

Does anyone of them have to be true?

It looks like there are pairs of them which are logically equivalent.

 

[Charles Link]

Te electric field E and electric potential V are two separate functions. Although the potential depends on the electric field, they are not proportional and the potential depends on the integral of the electric field over a path. The forum rules don't allow simply giving the answer, but the answer is quite simple. @SammyS The pairs are not logically equivalent. The equation the OP presents that E=V/d does have precise proportionality between E and V, but this equation is very misleading because it does not apply in general. It is for the special case of an ideal capacitor and E is the uniform electric field between the plates and V is the voltage drop across the plates. The equation really does not apply here, and the capacitor equation does not give V at any location between the plates where the E field is present. Although it looks like the right equation, it is totally irrelevant to this problem.

 

[SammyS]

Te electric field E and electric potential V are two separate functions. Although the potential depends on the electric field, they are not proportional and the potential depends on the integral of the electric field over a path. The forum rules don't allow simply giving the answer, but the answer is quite simple. @SammyS The pairs are not logically equivalent.

Are you saying that there is no pair that are not logically equivalent?

From a pure logic point of view.

(P implies Q) is logically equivalent to ((not Q) implies (not P)) .

It appears to me that we can find cases where one of these statements is the contrapositive of another.

 

[Charles Link]

Are you saying that there is no pair that are not logically equivalent?

From a pure logic point of view.

(P implies Q) is logically equivalent to ((not Q) implies (not P)) .

It appears to me that we can find cases where one of these statements is the contrapositive of another.

@SammyS It is somewhat difficult to answer your question without giving out the complete answer (at least what I am pretty certain is the correct answer), but none of the statements contains logical equivalence. To just give a counterexample for statement "d", a charged hollow conducting sphere has E=0 throughout the entire interior, but V is not equal to zero... editing.. And to give the OP something that might help them answer "b" and "c", what is the V and E for the point midway between two electrical charges of +Q and -Q? And I think the example I gave for "d" can also be used to answer "a".