Respuesta :
Question is incomplete.
Complete Question :
A particle with a charge of is +4.2 nC in a uniform electric field directed to the left. It is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is found to be: a. What work was done by the electric force? b. What is the potential of the starting point with respect to the endpoint? c. What is the magnitude of E?
Answer:
a) Work done by the electric force = W = 2.20 x 10^-6
b) Potential of the starting point with respect to end point = 523.80 V
c) Magnitude of E = 8.73 x 10^3 N/C
Explanation:
a) The charge has moved from point a to point b, hence the work done basically equals the change in kinetic energy.
Hence ,
W(a to b)= ΔK = Kb-Ka
W(a to b) = Kb
W(a to b) = 2.20 x 10^-6
b) W(a to b)/q = Va - Vb
Va - Vb = W(a to b) / q
=2.20 x 10^-6 J / 4.20 x 10^-9
=523.80 V
c) E = Va - Vb / l = 523.80 V / 0.060 = 8.73 x 10^3 N/C
a) Work done by the electric force should be [tex]W = 2.20 \times 10^-6[/tex]
b) Potential of the starting point with respect to end point should be 523.80 V.
c) Magnitude of E = [tex]8.73 \times 10^3 N/C[/tex].
Calculation of the work done, magnitude:
a)
Since The charge has moved from point a to point b, so the work done basically should be equivalent to the change in kinetic energy.
So,
W(a to b)= ΔK = Kb-Ka
W(a to b) = Kb
W(a to b) = [tex]2.20 \times 10^-6[/tex]
b)
Now
[tex]W(a\ to\ b)\div q = Va - Vb[/tex]
Va - Vb = [tex]W(a\ to\ b) \div q[/tex]
[tex]=2.20 \times 10^-6 J \div 4.20 \times 10^{-9}[/tex]
=523.80 V
c) [tex]E = Va - Vb \div l = 523.80 V \div 0.060[/tex]
=[tex]8.73 \times 10^3 N/C[/tex]
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