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A small sphere with mass 5.00*10-7kgand charge +7.00 m Cis released from rest at a distance ofabove a large horizontal insulating sheet of charge that has uniform surface charge density is σ=+8.00pC/m2. Using energy methods, calculate the speed of the sphere when it isabove the sheet.

Short Answer

Expert verified

The speed of the sphere when it is 0.100 m above the sheet is 1.45m/s.

Step by step solution

01

Step-1: Equation for finding the speed of the sphere

The sphere is having both mass and charge, thus is affected by the gravitational force as well as electrostatic force. The total mechanical energy will be conserved at the instant of dropping it from a height of la=0.400m, and when it reaches the height lb=0.100mabove the sheet, attaining speed.

Ulb+Tlb=Ulb+Tlb

At the height la=0.400m, the sphere is at rest, so, kinetic energy will be zero, Tla=0, but there will be gravitational potential energy and the electrostatic potential energy, which makes Ula=mgla+qV. But, the electric field is the gradient of potential, then E=Vdwhich givesV=Ed, in terms of distancela, V=Ela. Thus, localid="1668227458997" Ela=Ula=mgla+qEla, where Elais the total mechanical energy.

At the height , the sphere has attained kinetic energy also, then the total energy has components of gravitational potential energy, electrostatic potential energy, and kinetic energy, making Eb=Ub+Tb=mglb+qEEb+12mvb2

But electric field due to a plane sheet is given by E=σε0, where is the surface charge density, and is the permittivity =ε0=8.854×1012m3kg1s4A2.

Then, El=Ub=mgla+qσε0aandEb=Ub+Tb=mglb+qσε0lb+12mvb2

But as per the conservation law,

Ela=Elb

mglaa+qσε0la=mglb+qσε0lb+12mvb2

Solve for,

12mvb2=mgla+qσε0lamglb+qσε0lbvb=2mmgqσε0Ialb

02

Step-2: Calculation of the speed of sphere

At the final instance, lb=0.100mthe speedvb, is obtained by substituting the values of

m=5.00107kg,q=+7.00μC=+7106C,Ia=0.400m,lb=0.100mσ=+8.00pC/m2=+8.001012C/m2,ε0=8.854×1012m3kg1s4A2,g=9.8m/s2invb=2mmgqσε0IaIbvb=2510751079.87106810128.8541012(0.4000.100)vb=2510751079.8710688.8540.300vb=1.45m/s

Thus, the speed of the sphere at a distance 0.100m from the sheet is 1.45m/s.

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