Question

In: Physics

Two tiny particles that have charges +20.0 μC and -8.00 μC are separated by a distance...

Two tiny particles that have charges +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the magnitude and direction of the electric field halfway between these two charges? (k = 9.0 x 10 ^ 9 N x m ^ 2 / C ^ 2)

a. 25.2 x 10 ^ 6 N/C, directed at the negative charge

b. 25.2 x 10 ^ 6 N/C, directed at the positive charge

c. 25.2 x 10 ^ 5 N/C, directed at the negative charge

d. 25.2 x 10 ^ 5 N/C, directed at the positive charge

Expert Solution

Electric field at mid point due to 20C

E1=Kq/

K=9X

q=charge=20C=20XC

r=10cm=0.1m

E1=Kq/=9XX20X/

E1=18XN/C

Its direction is to right as q is positive (that is to the side of negative charge)

Electric field at mid point due to -8C

E2=Kq/

K=9X

q=charge=-8C=8XC (value)

r=10cm=0.1m

E2=Kq/=9XX8X/

E2=7.2XN/C

Its direction is also to right as q is negative charge (to the side of negative charge)

total field E=E1+E2=18X+7.2X=25.2X ( direction of E is to side of negative charge)

Answer is first option

genius_generous answered 2 years ago

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