Physics 260 Calculus Physics II: E&M Grist SOLUTION Worksheet One: Electrical Potential 1) When an electron moves from A to B along an electric field line the electric field does 3.94 1019 J of work on it. What are the electric potential differences: a) VB VA , b) VC VA , c) VC VB (a) VB – VA = ΔU/q = –W/(–e) = – (3.94 x 10–19 J)/(–1.60 x 10–19 C) = 2.46 V. (b) VC – VA = VB – VA = 2.46 V. (c) VC – VB = 0 (Since C and B are on the same equipotential line). 2) Consider a point charge q 1.0 C , point A at a distance d1 2.0m from q , and point B at a distance d 2 1.0m . a). If A and B are diametrically opposite each other, as in the figure (a), what is the electrical potential difference VA VB ? b). What is that electrical potential difference if A and B are located as in figure (b)? (a) The potential difference is VA VB q 4 0 rA 1 1 1.0 10 6 C 8.99 109 N m 2 C 2 4 0 rB 2.0 m 1.0 m q 4.5 103 V. (b) Since V(r) depends only on the magnitude of r , the result is unchanged. 3) A Gaussian sphere or radius 4.0 cm is centered on a ball that has a radius of 1.0 cm and a uniform charge distribution. The total (net) electric flux through the surface of the Gaussian sphere is 5.60 104 Nm 2 / C . What is the electric potential 12.0cm from the center of the ball? First using Gauss’ law, solve for the charge, q = εoΦ=+495.8 nC. Next solve for the potential: V q (8.99 109 N m 2 C2 )(4.958 107 C) 3.71104 V. 4 0 r 0.120 m Physics 260 Calculus Physics II: E&M Grist 4) The electrical potential in the space between two flat parallel plates 1 and 2 is given (in volts) by V 1500 x 2 , where x (m) is the perpendicular distance from plate 1. At x 1.0cm : What is the magnitude of the electric field and its direction relative to plate 1? Use the Del operation that I demonstrated in class. Since in this case the potential is a function of only one variable this becomes an ordinary derivative: dV E dx d ˆ 2 ˆ 2 ˆ ˆ ˆ i (1500 x )i ( 3000x)i ( 3000 V/m ) (0.0130 m)i (39 V/m)i. dx (a) The magnitude of the electric field is therefore E = 39 V/m. (b) The direction of E is î , or toward plate 1. 5) The electric potential at point in an xy plane is given by V ( 2.0V / m 2 )x 2 ( 3.0V / m 2 ) y 2 . What is the electric field at the point 3.0miˆ , 2.0mjˆ ? Again use the Del operation that I demonstrated in class: V (2.0V / m2 ) x 2 3.0V / m2 ) y 2 2(2.0V / m2 ) x; x x V E y ( x, y) (2.0V / m2 ) x 2 3.0V / m2 ) y 2 2(3.0V / m2 ) y . y y E x ( x, y) c c h h We evaluate at x = 3.0 m and y = 2.0 m to obtain E ( 12 V/m)iˆ (12 V/m)jˆ 6) a) What is the electric potential energy of two electrons separated by 2.0 nm? b) If the separation increases, does the potential; energy increase or decrease (a) Use the relationship for the potential energy of a pair of point charges with q1 = q2 = –e and r = 2.00 nm: U k 9 2 2 19 2 q1q2 e 2 8.99 10 N m C (1.60 10 C) k 1.15 1019 J. 2.00 109 m r r (b) Since U > 0 and U r–1 the potential energy U decreases as r increases.