answer all parts..show all work plz 3:52 PM webassign.net on Verizon LTF 7536 Th

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answer all parts..show all work plz

3:52 PM webassign.net on Verizon LTF 7536 The three charged pertices in the figure below are at the verticns c an lscaceles triangle twherd-2.5a cTaking4 8.30 uC, calculate the electric potential at point A, the midpoint of tha base Part 1 of 3 . Concapluai Since the midpeint of the base is closer to the two negertive charges. it pronably has negative pater a an the order of several kilovolts. Part 2 3-Calugorize Each charge creates a potential at the midpoint of the base. n crder to find the total electric potential at this point, we add the contributions from each of the charges The charged particla at the top vertex of the triange is a distence fram pairt A. Dy the Pythagcrean theorem, we have 5olving for the unknown distance, we obtain the foloning The distance at the charge to the left and the cistarce ., at the charge to the rignt frum pgirt A dre 'a-Q-d2. The potential at point A thn sunn the carr 1buitinns tom each charge at its distance from pairt A as follows (8.gg x 10g N m,2/D2' T x 10-6 c'( -)/m)

Explanation / Answer

Potential due to a point charge at a distance r from it is given as

V = kq / r

if q is -ve then V is -ve and if q is +ve then V also +ve.

distance of base charges from mid point,

r = 2.50/2 = 1.25 cm = 0.0125 m

so due to each charge , V1 = V2 = (9 x 10^9 x -8.30 x 10^-6) / 0.0125

V1 = V2 = - 5.976 x 10^6 Volt

distance of top charge, r = 0.0250 sin60 = 0.02165 m

V3 = (9 x 10^9 x 8.30 x 10^-6) / 0.02165

V3 = 3.45 x 10^6 volt.

Vnet = V1 + V2 + V3 = [ - 5.976 - 5.976 + 3.45 ] x 10^6

      = - 8.50 x 10^6 Volt

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