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Standard XII

Physics

Coulomb's Law - Mathematical Form

Two positive ...

Question

Two positive charges $q_{1} = 20 C$ and $q_{2} = 6 C$ are separated by diameter of $3 m$ . Then what is the force produced by them?

80 N

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53.3 N

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56 N

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72.5 N

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Solution

The correct option is D $53.3 N$

Here, $r = \frac{3}{2} = 1.5 m$ and charges are $q_{1} = 20$ and $q_{2} = 6$

Coulomb's law,

$F = k \frac{q_{1} q_{2}}{r^{2}}$

Therefore, (for simplicity assume k=1)

$F = \frac{20 \times 6}{(1.5)^{2}} = 53.3 N$

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Two positive charges q1=20C and q2=6C are separated by diameter of 3m. Then what is the force produced by them?

The correct option is D 53.3N Here, r=32=1.5m and charges are q1=20 and q2=6 Coulomb's law, F=kq1q2r2 Therefore, (for simplicity assume k=1) F=20×6(1.5)2=53.3N

Two positive charges $q_{1} = 20 C$ and $q_{2} = 6 C$ are separated by diameter of $3 m$ . Then what is the force produced by them?

The correct option is D $53.3 N$

Here, $r = \frac{3}{2} = 1.5 m$ and charges are $q_{1} = 20$ and $q_{2} = 6$

Coulomb's law,

$F = k \frac{q_{1} q_{2}}{r^{2}}$

Therefore, (for simplicity assume k=1)

$F = \frac{20 \times 6}{(1.5)^{2}} = 53.3 N$