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, r_B = 2 r_A

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

E_A / E_B = (1 / r_A² ) / (1 / r_B²) { eq. 2 }

E_A / E_B = r_B² / r_A²

E_A / E_B = (r_B / r_A)² { eq. 3 }

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

E_A / E_B = [(2 r_A) / r_A]²

E_A / E_B = 4 r_A² / r_A²

E_A / E_B = 4 / 1

genius_generous answered 5 days ago

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