let us specify the location as a distance x fromq1
distance between two charges d = 20 cm
q 1 = 1.90μC = 3 *10 -6 C
q 2 = 4.45μC = 7 *10 -6 C
setting the magnitudes of the individual fields equaland solving for x
E1= E2
k q1 /x2 = k q2/ ( x+d ) 2
1 / x2= (q2 /q1 ) / ( x + d ) 2
q 2 / q1 = 4.45/ 1.90 =2.34
1 / x2 =2.34 / ( x + d )2
1 / x=(√2.34) / x+ d
x+ d = 1.53 x
d= 0.53 x
x= 5.5*10-2/ 0.53
= 0.0359 m
= 3.59 cm
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