Electric Potential: Derive Expression for Vxy

[thereddevils]

Homework Statement

Derive an expression for p.d , V between point X and point Y in an electric field due to a single point charge +Q as shown in the figure attached .

Homework Equations

The Attempt at a Solution

Vxy=Vy-Vx

=\frac{Q}{4\pi\epsilon_o r_y}-\frac{Q}{4\pi\epsilon_o r_x}

=\frac{Q}{4\pi\epsilon_o}(\frac{1}{r_y}-\frac{1}{r_x})

I guess it's not that simple , because this question carries 5 marks .

sorry , i am not sure which is the icon to wrap these math codes .. could someone pls modify my post ?

 

[Doc Al]

I guess it's not that simple , because this question carries 5 marks .

Looks OK to me. Were you supposed to derive it from the electric field?

sorry , i am not sure which is the icon to wrap these math codes .. could someone pls modify my post ?

To use Latex, wrap it with tex tags:
[tag]x^2[/tag] = x^2

replace 'tag' with 'tex'

 

[thereddevils]

Looks OK to me. Were you supposed to derive it from the electric field?


To use Latex, wrap it with tex tags:
[tag]x^2[/tag] = x^2

replace 'tag' with 'tex'

thanks , but this is another way of doing it ..

Vxy=Vy-Vx

and from work definition , dW=F dr

V_{xy}=\frac{F dr}{q}-\frac{F dr}{q}

=\frac{1}{q}\int^{r_y}_{\infty} F dr-\frac{1}{q}\int^{r_x}_{\infty} F dr

=\frac{1}{q}\int^{r_y}_{r_x} \frac{Qq}{4\pi \epsilon_o r^2}

=\frac{Q}{4\pi \epsilon_o }[-\frac{1}{r}]^{r_y}_{r_x}

=\frac{Q}{4\pi \epsilon_o }(\frac{1}{r_x}-\frac{1}{r_y})

When it's derived from the definition of electric potentials , the final products are different , or can i say since electric potential is a scalar , so that shouldn't matter ?

 

[Doc Al]

When it's derived from the definition of electric potentials , the final products are different , or can i say since electric potential is a scalar , so that shouldn't matter ?

You need to calculate the work done per unit charge against the electric force, so you're missing a minus sign in the expression for force. (Signs matter!)

 

[thereddevils]

You need to calculate the work done per unit charge against the electric force, so you're missing a minus sign in the expression for force. (Signs matter!)

oh , thank you very much sir !