Consider two point charged objects separated by a distance of 5.20 cm. The charg

Question

Consider two point charged objects separated by a distance of 5.20 cm. The charge on the left hand object is +43.5 C. The charge on the right hand object is -43.5 C.

What is the strength (magnitude) of the electric field due to the left hand object only at the location exactly halfway in between the two charged objects?

What is the strength (magnitude) of the electric field due to the right hand object only at the location exactly halfway between the two charged objects?

What is the strength of the net electric field due to the BOTH objects at the point halfway between the two charged objects?

What would be the force on a proton placed at the point halfway between the objects?

Explanation / Answer

Ans:-

Given data, distance = 5.20cm r = 2.6cm = 2.6*10^-2m

ql = 43.5C qr = -43.5c

Electric field equation

E = (1/40) *(|ql|*/r^2)

1 E1 = 9*10^9*43.5*10^-6/(2.6*10^-2)^2

=579142.01N/C

2 E2 = 579142.01 N/C = 5.8*10^6N/C

3 Enet = E1 +E2 = 1158284.024 = 1.16*10^6N/C

4 charge of proton = +1.6*10^-19C

E = F/q

F = E *q = 1.16*10^6*1.6*10^-19 = 1.86*10^-13N

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