Assume a spherically symmetric electric field whose value is plotted against distance from the origin in the figure below. k,Q, R and a(alpha) are constants. assume the potential at infinity V=0 the potential at r=0 is found to be (-kQ)/(3R).

What is the value of a?

I am lost as to how to even start this.....I know that V=-integral E dr, but not totally sure if that even helps.

Question

Assume a spherically symmetric electric field whose value is plotted against distance from the origin in the figure below. k,Q, R and a(alpha) are constants. assume the potential at infinity V=0 the potential at r=0 is found to be (-kQ)/(3R).  

What is the value of a?

I am lost as to how to even start this.....I know that V=-integral E dr, but not totally sure if that even helps.

anonymous · Answer

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anonymous · Answer

Yes, V = - integral E dr is exactly what you need, just start out at infinity and integrate in towards the origin in sections, using the appropriate form of E in each region.
I expect that will give you an expression for alpha

anonymous · Answer

so take $$-\int\limits_{3R}^{\infty} -\alpha*kQ/(r^2) dr$$  $$-\int\limits_{R}^{2R}kQ/r^2 dr$$ and $$-(1/2)*R*(kQ/R^2)$$ and then do what with them?  Set the sum of them equal to the total change in potential?

anonymous · Answer

remember that you're integrating in towards the origin from infinity
the potential at infinity is set to be zero, so that means the integral of -E from infinity to r simply is the potential at that value of r
so you're going to end up with an expression for V at r=0 and just equate that to the value given in the question
then you can solve for alpha

anonymous · Answer

it seems to me your expression is correct except that you have -V instead of V for some reason

anonymous · Answer

Thanks that makes sense now.   Where are you talking about the -V coming from?

anonymous · Answer

it's easy to get the signs wrong in this sort of question
you have the limits on your integrals the wrong way round, hence all your terms are wrong by a factor of -1

anonymous · Answer

same goes for the last term that you already integrated

anonymous · Answer

Ah, I see what you are talking about.  I was going from 0 to infinity instead of infinity to zero.

anonymous · Answer

yeah, right

anonymous · Answer

just let me know what you get for alpha when you're done

anonymous · Answer

I got alpha = 1, but I'm pretty sure that is wrong...

anonymous · Answer

$$-(\alpha*kQ/(3R))-kQ/(2R)+kQ/(2R)=-kQ/(3R)$$  I'm pretty sure that there is a sign error somewhere, but I am not completely sure where.

anonymous · Answer

If my working is correct, I think your middle term on the left should be positive, the rest agrees with my expression.

anonymous · Answer

Okay, I see where I forgot to make the second integral negative.  Now I get alpha to be equal to 4