Question

In: Physics

The diagram shows an isolated, positive charge Q, where point B is twice as far away...

The diagram shows an isolated, positive
charge Q, where point B is twice as far away
from Q as point A.
+Q ____A____ B
0 10cm 20cm
What is the ratio of the electric field
strength at point A to the electric field
strength at point B?

Solutions

Expert Solution

electric field strength is inverseley proportional to the square of distance :

E 1 / r2                                { eq. 1 }

suppose, rB = 2 rA

the ratio of the electric field strength at point A to the electric field strength at point B is given as :

EA / EB = (1 / rA2 ) / (1 / rB2)                            { eq. 2 }

EA / EB = rB2 / rA2

EA / EB = (rB / rA)2                                    { eq. 3 }

inserting the value of 'rB' in eq.3,

EA / EB = [(2 rA) / rA]2

EA / EB = 4 rA2 / rA2

EA / EB = 4 / 1


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