Item 2 A point charge of 3.00 C is located in the center of a spherical cavity o

Question

Item 2 A point charge of 3.00 C is located in the center of a spherical cavity of radius 6.50 cm inside an insulating spherical charged solid. The charge density in the solid is 7.35 × 10-4 C/m3. Part A Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express your answer with the appropriate units. EValue Units Submit Give Up Part B Find the direction of this electric field radially outward radially inward Submit Give Up Continue >

Explanation / Answer

A point charge of -3 uC is located in the center of a spherical cavity of radius 6.5 cm inside an insulating spherical charged solid

field due to point charge is E = k*q/r1^2 = (9*10^9*3*10^-6)/(9.5*10^-2)^2 = 3*10^6 N/C

the charge associated with the volume is Q = rho*V

given that rho= 7.35*10^-4 C/m^3

V = (4/3)*3.142*(9.5^3-6.5^3)*10^-6 = 2.44*10^-3 m^3

therefore Q = rho*V = 7.35*10^-4*2.44*10^-3 = 1.794*10^-6 C

so electric field at the given point is E2 = k*Q2/r2^2 = (9*10^9*1.794*10^-6)/(9.5*10^-2)^2 = 1.789*10^6 N/C

Net electric field is Enet = E1+E2 = (-3*10^6)+(1.789*10^6) = -1.211*10^6 N/C

magnitude is E = 1.211*10^6 N/C

negative sign showns that net field is directed inwards

so the answer is radially inward

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