As shown in the figure below, a uniform electric field has magnitude E = 230 N/C

Question

As shown in the figure below, a uniform electric field has magnitude E = 230 N/C and is directed to the right. A particle with charge +3.7 nC moves along the straight line from a to b.

magnitude  
As shown in the figure below, a uniform electric field has magnitude E = 230 N/C and is directed to the right. A particle with charge +3.7 nC moves along the straight line from a to b. What is the electric force that acts on the particle? What is the work done on the particle by the electric field? What is the potential difference Va? Vb between points a and b?

Explanation / Answer

a). electric force= qE

= 230*3.7*10^-9

= 8.57*10^-7 N

=0.857uN to the right

b).work done= Fd

= 8.57*10^-7*.25

=0.21275 uJ

c).Va-Vb= E.r

= 57.5 V

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