Multiple-Choice Homework Problem 2.5 The figure to the right shows two charged p

Question

Multiple-Choice Homework Problem 2.5 The figure to the right shows two charged particles: an electron labeled e and a proton labeled p. The two particles are a distance d apart and the electron is a distance d from the point a. What is the direction of the electric field at the point a? Select One of the Following: (a) The electric field is zero (b) The electric field points in the positive z direction. (c) The electric field points in the negative r direction. (d) The electric field points in the positive y direction. (e) The electric field points in the negative y direction. Multiple-Choice Homework Problem 2.6 The figure to the right shows four systems of charged particles labeled (a)-(d). Each system contains two charges. Systems (a) and (c) contain two (a) +q. positively charged particles with charge +q. Systems (b) and (d) contain particles with equal but opposite charge ta. The particles (b) +q. are a distance d from the central point P in systems (c) and (d and a distance 2d from the central point in (a) and (b). Rank the magnitude of the electric field, E, at the point P of the four (c) systems. (d) Select One of the Following: (a) El Ed Ea E (b) E Eb Ec Ed (c) E Ee El (d) E. Ec Eb Ed

Explanation / Answer

a) charge of proton=q=1.6*10^(-19) C

charge of electron=-q

for any charged particle with charge Q, electric field at a distance of d is given as

9*10^9*Q/d^2

so here electric field at a due to electron=-9*10^9*q/d^2

hence magnitude=9*10^9*q/d^2, and direction is along +ve x axis.

and electric field at a due to proton=9*10^9*q/(2d)^2=2.25*10^9*q/d^2

hence magnitude =2.25*10^9*q/d^2 and direction is along -ve x axis.

hence net field=-6.75*10^9*q/d^2

and direction will be along +ve x axis.

hence, option b is correct.

problem 2.6:

as we know,

electric field due to a positive charge is away from the charge and due to a negative charge is towards the chrge.

electric field of a charge Q at a distance d is given by

k*Q/d^2

where k=9*10^9

a) electric field due to left side +q =k*q/(2d)^2=k*q/(4*d^2), towards +ve x axis.

electric field due to right side +q =k*q/(2d)^2=k*q/(4*d^2), towards -ve x axis.

hence net field=0

b)electric field due to left side +q =k*q/(2d)^2=k*q/(4*d^2), towards +ve x axis.

electric field due to right side -q =k*q/(2d)^2=k*q/(4*d^2), towards +ve x axis.

hence net field=2*k*q/(4*d^2)=k*q/(2*d^2), towards +ve x axis.

c) electric field due to left side +q =k*q/(d)^2, towards +ve x axis.

electric field due to right side +q =k*q/(d)^2, towards -ve x axis.

hence net field=0

d) electric field due to left side +q =k*q/(d)^2, towards +ve x axis.

electric field due to right side -q =k*q/(d)^2, towards +ve x axis.

hence net field=2*k*q/d^2, towards +ve x axis.

so ranking the electric field in ascending order,

we get Ea=Ec<Eb<Ed

hence option d is correct.

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