# physics

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The drawing shows a square, each side of which has a length of L = 0.25 m. Two different positive charges q1 and q2 are fixed at the corners of the square. Find the electric potential energy of a third charge q3 = -4.35 10-8 C placed at corner A and then at corner B.

• physics -

here are the other charges
q1= +1.5e-9
q2=+4.0e10-9

• physics -

Electric potential is a scalar, so you add the potential from each of the other charges.

Vtotal=v1+V2=k(q1/distance+q2/distance2)

Now, for the PE, EPE=Vtotal*q3

• physics -

but i need two ansers for the epe of charge 3 at point a and b. do I just multiply the vtotal by q3 then multiply each charge by that to get the numer for pt a and b

• physics -

diagonal of the square = 0.25•sqrt2=0.35 m.

point A (top, right)
φ1=k•q1/0.35 =9•10^9•1.5•10^-9/0.35 =38.6 V
φ2=k•q2/0.25 =9•10^9•4•10^-9/0.25 =144 V
φ = φ1+ φ2 = 182.6 V.
PE=q• φ =-4.35•10^-8• 182.6 = -7.94•10^-6 J.
point B (top, left)
φ1=k•q1/0.25 =9•10^9•1.5•10^-9/0.25 =54 V
φ2=k•q2/0.35 =9•10^9•4•10^-9/0.35 =102.9 V
φ = φ1+ φ2 = 156.9 V.
PE=q• φ =-4.35•10^-8• 156.9 = -6.82•10^-6 J.