The three charged particles in the figure below are at the vertices of an isosceles triangle (where d = 5.00 cm). Taking q = 7.60 µC, calculate the electric potential at point A, the midpoint of the base. 2d
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The three charged particles in the figure below are at the vertices of an isosceles triangle (where d = 5.00 cm). Taking q = 7.60 µC, calculate the electric potential at point A, the midpoint of the base.
The electric potential at point A, the midpoint of the base.
Given:
The magnitude of all the charges is
The distance between two negative charges is .
The distance between negative and positive charges is .
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- A charge Q=-13 nC is placed at the origin. What is the electric potential at point A=(4.2, -4.7) cm? Provide your answer in volts to the nearest integer.The three charges in the figure below are at the vertices of an isosceles triangle. Let = 8.00 nC and calculate the electric potential at the midpoint of the base. (Let d, = 1.50 cm and d, = 7.00 cm. d2 d1 kVFind a value (in terms of k.) for AV between locations x, (i.e. 9 grid points from the q=+1 nC charge) and x3 (i.e. 3 grid points from the q=+1 nC charge). Each grid space is 0.5 m and 1 nC= 1.0 x 10° C. (Hint: you may want to use the formula for the potential around a point charge derived in the mini-lecture.) Show all work.
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- The picture below shows two concentric rings. The outer ring has a charge of -300nC and a radius of 18mm. The inner ring has a linear charge density of 12uC/m and a radius of 9mm. What is the electric potential at point P, which is 30mm above both rings.Chapter 25 Q178: Three circular, nonconducting arcs of radius R = 3 cm.The charges on the arcs are q1 = 5 pC, q2 = 10 pC , q3 = 15 pC. With V = 0 at infinity, what is the net electric potential of the arcs at the common center of 45.0/ 45.0° curvature? R 93 92Two rods of equal length L=1.18 m form a symmetric cross. The horizontal rod has a charge of Q=300 C, and the vertical rod has a charge of −Q=−300 C. The charges on both rods are distributed uniformly along their length. Calculate the potential at the point P at a distance D=2.63 m from one end of the horizontal rod. Special note: You're going to have to take the difference between the potentials made by the positive and negative rods, which are similar numbers. That means that to get an accurate answer you need to keep more digits than usual while you're doing your calculation.
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