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Punchlinegirl
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Three charges are located at the vertices of an isosceles triangle. A positive charge, of 5.7 x 10^-9 C is located at the top, and two negative charges each of 5.4 x 10^-9 C are located at the bottom. The lengths of the 2 sides are 7.1 cm and the length of the base is 1.3 cm. Calculate the electric potential at the midpoint of the base. Answer in units of V.
I used the principle of superposition. I found that the length from the charge to the midpoint was .0065 m and the length of the other side when the triangle was cut in half was .0707 m.
So V= 9 x 10^9 [ (5.7 x 10^-9/.0707) + (5.4 x 10^-9 /.0065) + (5.4 x 10^-9/.0065)
So V= 1.57 x 10^4, but this isn't right... can someone please tell me what I'm doing wrong? Thanks in advance.
I used the principle of superposition. I found that the length from the charge to the midpoint was .0065 m and the length of the other side when the triangle was cut in half was .0707 m.
So V= 9 x 10^9 [ (5.7 x 10^-9/.0707) + (5.4 x 10^-9 /.0065) + (5.4 x 10^-9/.0065)
So V= 1.57 x 10^4, but this isn't right... can someone please tell me what I'm doing wrong? Thanks in advance.