1. A ball of aluminum foil mass 5.20 x 10^-2 kg isplaced in an electric field di
ID: 1725176 • Letter: 1
Question
1. A ball of aluminum foil mass 5.20 x 10^-2 kg isplaced in an electric field directed vertically upward. ifthe charge on the foil is 7.0 uC. find the strengh of thefield. What will cause the ball to "float" without fallingdue to the earth gravity field. 2. Find the lectric potential .250 um from an electron(qe=1.602x10^-19C). What is the potential difference betweentwo points .350 and .375 cm from an electron? 3. A potential difference of 9.0V is found to produce acurrent .50 A in a 75.0 m lenght of wire a uniform diameter of 2.85m. what is the resistance of the wire. What is theresistivity of the wire? 1. A ball of aluminum foil mass 5.20 x 10^-2 kg isplaced in an electric field directed vertically upward. ifthe charge on the foil is 7.0 uC. find the strengh of thefield. What will cause the ball to "float" without fallingdue to the earth gravity field. 2. Find the lectric potential .250 um from an electron(qe=1.602x10^-19C). What is the potential difference betweentwo points .350 and .375 cm from an electron? 3. A potential difference of 9.0V is found to produce acurrent .50 A in a 75.0 m lenght of wire a uniform diameter of 2.85m. what is the resistance of the wire. What is theresistivity of the wire?Explanation / Answer
1) The electric force is F =Eq but F = mg The strength of the field is E = (mg)/q Here m = 5.20 x 10^-2 kg q =7.0*10-6C 2) We have the formula for the electric potential is givenby V = kq/r2 Here q = -1.60*10-19C k =8.90*10^9 N.m2/C2 r = 0.250*10-6m The potential difference is V = kq/r2 -kq/r'2 Here r = 0.350*10-2m r' =0.375*10-2m Substitute the values. 3) The resistance of the wire is R = V/I = (9.0 V)/(0.50 A) The resistance of the wire is given by R = L/(r2) The resistivity of the wire is = (R(r2) )/L Here r = 1.425 m L = 75.0 m Substitute the values.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.