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Nata [24]
3 years ago
8

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Physics
3 answers:
garri49 [273]3 years ago
8 0

Solution:

The force between two charges given by Columb's law ithe force between two charges given by Coloumb law is\\
\left | F \right |=\frac{kq_{1}q_{2}}{r^{2}}\\\\
where k is the constant equal to\\  \rn Now,\\
We have \\r=0.008mand q1=q2=-25.0\times10^{-9}\\
\left | F \right |=\left ( 8.099\times10^{9}\frac{Nm^{2}}{c^{2}} \right )\times\left ( \frac{25.0\times10^{-9}\times 25.0\times10^{-9}}{0.008^{2}} \right )\\
=1.21\times10^{-3}

Daniel [21]3 years ago
7 0

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

Here, q_{1} andq_{2} are the charges on the pith balls, r is the separation between the charges and k is constant and its value is  8.99\times 10^9 N m^2/C^2.

Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

valina [46]3 years ago
7 0

Solution:

The force between two charges given by Columb's law ithe force between two charges given by Coloumb law is\\
\left | F \right |=\frac{kq_{1}q_{2}}{r^{2}}\\\\
where k is the constant equal to\\  \rn Now,\\
We have \\r=0.008mand q1=q2=-25.0\times10^{-9}\\
\left | F \right |=\left ( 8.099\times10^{9}\frac{Nm^{2}}{c^{2}} \right )\times\left ( \frac{25.0\times10^{-9}\times 25.0\times10^{-9}}{0.008^{2}} \right )\\
=1.21\times10^{-3}

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