ID: B1UCSD Physics 2B Unit Exam 1B Charges, Force & FieldsAnswer SectionMULTIPLE CHOICE1. ANS: BMake a ratio of the two forces using the fact that the only change is that the new distance is rnew=1.65 *rFnewF=kq1q2rnewÊËÁˆ¯˜2kq1q2r( )2=rrnewÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜2⇒Fnew=FrrnewÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜2=22N()11.65ÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜2=8.08NTOP: COULOMB FORCE CONCEPT2. ANS: DThe electric field from a point charge is E=kQr2. Outside a charged spherical shell, the field looks the same as if the whole charge were located in a point at the center. Halfway between the shells, the distance from their center is 32R. Hence the field isE=kQr2=kQ32RÊËÁÁÁÁˆ¯˜˜˜˜2=49kQRThe other way to do this is to construct a Gaussian Surface at r=3R/ 2. The enclosed charge is +Qand the surface area A=4πr2=4π32RÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜2=9πR2. The electric field isE=Qencε0A=+Q4π ε094R2ÊËÁÁÁÁÁÁˆ¯˜˜˜˜˜˜=49kQR2where we used k=14πε0TOP: ELECTRIC FIELD SPHERICAL3. ANS: DFundamental properties of conductors:Electric field is zero insideElectric field is perpendicular at surfaceExcess charge must reside at surface onlyTOP: CONDUCTOR FIELD CONCEPT
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