Two point charges of +16 μC and – 9 μC are placed 8 cm apart in air. At what point on the line joining the charges is the electric field intensity zero?
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q/r² = q'/r'²
16×10^-6/x² = 9×10^-6/(0.08-x)²
16/x² = 9/(0.08-x)²
16(0.08-x)² = 9x²
0.32-4x = 3x
7x = 0.32
x = 0.32/7
x = 0.0457m
or x = 4.6 cm
16×10^-6/x² = 9×10^-6/(0.08-x)²
16/x² = 9/(0.08-x)²
16(0.08-x)² = 9x²
0.32-4x = 3x
7x = 0.32
x = 0.32/7
x = 0.0457m
or x = 4.6 cm
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