Question

The following scenarios will ask you to compare the electric field at point P for four differnet situations. Situation B: Situation D: 20) ·P 1/2R Situation A: Situation C: 2Q 1/2R How does the magnitude of the electric field at P compare for Situations A and B? EA EB EA-2EB AEB O EA EB EA-4EB How does the magnitude of the electric field at P compare for Situations A and C? EA=Ec EA 2E EA E 4 TC How does the magnitude of the electric field at P compare for Situations A and D? EA-8ED EA-4ED EA2ED

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Answer #1

Electric field due to a point charge at distance 'r' is given by:

E = k*q/r^2

where direction of electric field is towards the negative charge and away from positive charge.

Part A.

Electric field at P for situations A and B

At point A:

EA = k*QA/rA2

Given that, QA = Q, and rA = R, So

EA = k*Q/R2

At point B:

EB = k*QB/rB2

Given that, QB = Q, and rB = R/2, So

EB = k*Q/(R/2)2 = 4*k*Q/R2

EB = 4*EA

EA = (1/4)*EB

Correct option is A.  

Part B.

Electric field at P for situations A and C

At point A:

EA = k*QA/rA2

Given that, QA = Q, and rA = R, So

EA = k*Q/R2

At point C:

EC = k*QC/rC2

Given that, QC = 2Q, and rC = R, So

EC = k*2Q/(R)2 = 2*k*Q/R2

EC = 2*EA

EA = (1/2)*EC

Correct option is D.  

Part C.

Electric field at P for situations A and D

At point A:

EA = k*QA/rA2

Given that, QA = Q, and rA = R, So

EA = k*Q/R2

At point D:

ED = k*QD/rD2

Given that, QD = 2Q, and rD = R/2, So

ED = k*2Q/(R/2)2 = 8*k*Q/R2

ED = 8*EA

EA = (1/8)*ED

Correct option is D.  

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