Consider the following figure. (If you need to use co or -oo, enter INFINITY or

Question

Consider the following figure. (If you need to use co or -oo, enter INFINITY or -INFINITY, respectively.) 10 x (cm) 5 24 x (cm) -q 10 -2q +q +3q Find the total electric field in N/C at x 5.00 cm in part (b) of the figure above given that q = 5.00 nC. (a) N/C (b) Find the total electric field in N/C at x = 11.30 cm in part (b) of the figure above. (Include the sign of the value in your answer.) N/C If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc., and what will its value(s) be? Use the following as necessary: q.) (c)

Explanation / Answer

part a:

field due to negative charge is towards the charge.

field due to positive charge is away from the charge.

electric field due to -2q at x=5 cm:

as charge is -ve, field is along -ve x axis.

distance=5-1=4 cm=0.04 m

field magnitude=k*charge/distance^2

=9*10^9*(2*5*10^(-9))/0.04^2

=56250 N/C

field due to +3q:

as charge is positive, field is away from the charge and along -ve x axis.

distance=3 cm=0.03 m

field magnitude=k*charge/distance^2

=9*10^9*(3*5*10^(-9))/0.03^2

=1.5*10^5 N/C

field due to -q:

as charge is negative, field is along +ve x axis.

distance=14-5=9 cm=0.09 m

field magnitude=k*q/distance^2

=9*10^9*(5*10^(-9))/0.09^2

=5555.6 N/C

so total field is along -ve x axis.

field magnitude=56250+1.5*10^5-5555.6=2.007*10^5 N/C

so answer=-2.007*10^5 N/C

part b:

electric field due to -2q at x=11.3 cm:

as charge is -ve, field is along -ve x axis.

distance=11.3-1=10.3 cm=0.103 m

field magnitude=k*charge/distance^2

=9*10^9*(2*5*10^(-9))/0.103^2

=8483.4 N/C

field due to +q:

as charge is positive, field is towards +ve x axis.

distance=11.3-5=6.3 cm=0.063 m

field magnitude=9*10^9*5*10^(-9)/0.063^2

=1.1338*0^4 N/C

field due to +3q:

as charge is positive, field is away from the charge and along +ve x axis.

distance=11.3-8=3.3 cm=0.033 m

field magnitude=k*charge/distance^2

=9*10^9*(3*5*10^(-9))/0.033^2

=1.2397*10^5 N/C

field due to -q:

as charge is negative, field is along +ve x axis.

distance=14-11.3=2.7 cm=0.027 m

field magnitude=k*q/distance^2

=9*10^9*(5*10^(-9))/0.027^2

=6.1782*10^4 N/C

total field is along +ve x axis.

net field magnitude=1.8861*10^5 N/C

part c:

final charge will be all at infinity as that will reduce the total potential energy.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.