The point-like charges carrying charges of $+ 3 \times 10^{- 9} C$ and $- 5 \times 10^{- 9} C$ are $2 m$ apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

Open in App

Solution

We use coulomb's law to calculate the magnitude of the force.
$F = k \frac{q_{1} q_{2}}{r^{2}}$

$q_{1} = + 3 \times 10^{- 9} C$
$q_{2} = - 5 \times 10^{- 9} C$
$r = 2 m$
$k = 9 \times 10^{9} N m^{2} C^{- 2}$
Therefore,
$F = \frac{9 \times 10^{9} \times 3 \times 10^{- 9} \times 5 \times 10^{- 9}}{4}$

$F = 3.37 \times 10^{- 8} N$
Since, the point charges have opposite signs, the force will be attractive.

Suggest Corrections

Similar questions

Q. Two identical metal sphere carrying charges of

9 \times 10^{- 9} C

and

- 3 \times 10^{- 9}

C are brought in contact and then they are separated by a distance of 3 m and kept in a medium of dielectric constant 5. what is the electrostatic force between them ?

Q. Two point charges

+ 3 μ C

and

- 3 μ C

are kept

20 c m

apart in air. Find the electric field midway between the charges on the line joining them. What will be the force experienced by a charge

+ 1.5 \times 10^{- 9} C

placed at this point ?

Two charges $q_{1} = 12 \times 10^{- 9} C$ and $q_{2} = - 12 \times 10^{- 9} C$ are placed $10 c m$ apart. The potential at a point $6 c m$ from $q_{1}$ on the line joining the two charges is :

Two point charges $Q_{1} = 8 \times 10^{- 9} C$ and $Q_{2} = - 6 \times 10^{- 9} C$ are separated by $10$ cm in air. The intensity of electric field at the midpoint between the point charges is :

Q. Two spherical bodies, each of mass

50

kg, are placed at a separation of

20

cm, Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude, The magnitude of the charge placed on either body is

x \times 10^{- 9} C

. Find the value of

x

Join BYJU'S Learning Program

The point-like charges carrying charges of +3×10−9C and −5×10−9C are 2 m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

We use coulomb's law to calculate the magnitude of the force.F=kq1q2r2q1=+3×10−9 Cq2=−5×10−9 Cr=2 mk=9×109 N m2 C−2Therefore,F=9×109×3×10−9×5×10−94F=3.37×10−8 NSince, the point charges have opposite signs, the force will be attractive.

The point-like charges carrying charges of $+ 3 \times 10^{- 9} C$ and $- 5 \times 10^{- 9} C$ are $2 m$ apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.