If the rod has a total charge of 6.27 C, find the horizontal component of the el

Question

If the rod has a total charge of 6.27 C, find the horizontal component of the electric field at O, the center of the semicircle. The value of the Coulomb constant is 8.98755×10^9 N m^2/C^2

.Define right as positive

b) Determine the value of the electric potential V at the center of the semicircle

I need help with part b please explain step by step with the solution. Thank You

-6.27 uC If the rod has a total charge of-6.27 ,C, find the horizontal component of the elec- tric field at O, the center of the semicir- cle. The value of the Coulomb constant is 8.98755 × 109 N·m2/C2 . Define right as pos- itive Answer in units of N/C. 009 (part 2 of 2) 10.0 points Determine the value of the electric potential V at the center of the semicircle Answer in units of J/C.

Explanation / Answer

Hi,

Well in this case you must remember know two thing:

1) You are dealing with a continuous charge distribution, so you will have to use some calculus.

2) The electric potential of a sole element of charge.

So, for starters we imagine a punctual charge on the semicircle, a point with no particular properties (such at being at the ends of the semicircle); and we imagine the electric potential that the element causes at the center. Translating that into a equation we have the following:

dV = k dq/r ; where V is the electric potential, q is the charge in that point, k is the electric constant and r is the distance from that point to the center of the semicircle.

Now, to find the electric potential that the semicircle causes at its center we only have to add all the potentials that any point on the semicircle creates over that point. Of course, to do that we use an integral:

V = k dq/r ; in this integral k is a constant, so can be put outside of it. The distance r is a constant because any point on the semicircle maintains the same distance to the center (which is the radius of the semicircle), therefore it can be put outside of the integral.

So, at end, we musts solve this integral : V = (k/r) dq , which is easy as the result is simply dq = Q ; which is the total charge of the semicircle.

So, the only thing we must find is the radius to solve this problem, and as they are giving us the lengh of the semicircle we can find is easyly doing the following:

L = 2*r ; where L is the length of the circle and r is the radius. But as it is a semicircle, the equation should be:

L = *r ; so ; r = L/ = (0.165 m) / = 0.053 m

Finally, the electric potential is: V = (9*109 N*m2/C2)*(6.27 C)/(0.053 m) = 1.07*1012 N*m/C = 1.07*1012 J/C

Note: in this case I used the value of 6.27 C, if I had used the value of -6.27*10-6 C the result would have been

V = -1.07*106 J/C

I hope it helps.

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