What is the value of r given the change in electric potential?

[mrcotton]

The electric potential at a distance r from a positive point charge is 45V. The potential
increases to 50 V when the distance from the charge decreases by 1.5 m. What is the
value of r?

A 1.3m
B 1.5m
C 7.9m
D 15m

Homework Statement

the answer is D

Homework Equations

V=(1/4∏ε) q/r
v proportional to 1/r

The Attempt at a Solution

5=(q/(1/4∏ε))*((1/r-1.5)-(1/r))

How embarrassing I just can’t get my head around this one. Am I setting the equation up correct to solve.
If I have I need desperate help with the algebra.
Thank you

 

[SammyS]

The electric potential at a distance r from a positive point charge is 45V. The potential
increases to 50 V when the distance from the charge decreases by 1.5 m. What is the
value of r?

A 1.3m
B 1.5m
C 7.9m
D 15m

Homework Statement

the answer is D

Homework Equations

V=(1/4∏ε) q/r
v proportional to 1/r

The Attempt at a Solution

5=(q/(1/4∏ε))*((1/r-1.5)-(1/r))

How embarrassing I just can’t get my head around this one. Am I setting the equation up correct to solve.
If I have I need desperate help with the algebra.
Thank you

You have one equation and two unknowns.

Taking the ratio $\displaystyle \ \ \frac{V(r)}{V(r-1.5)}\ \$ might be more helpful.

Otherwise, use your equation along with $\displaystyle \ \ V(r)=45=\frac{q}{4\pi\epsilon_0\,r}\ \$ then eliminate q & solve for r.

 

[gneill]

You're going to want to get rid of the constants you don't have values for (like the charge value q) so you'll probably want to set it up as a ratio so that they'll cancel out.

EDIT: SammyS got their first!

 

[mrcotton]

Thank you guys.
I did it like this with your help.
Is this mathematicaly sound.
Thanks
Mr C

 

[SammyS]

Thank you guys.
I did it like this with your help.
Is this mathematically sound.
Thanks
Mr C

That answer is correct.