In: Physics
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?
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