Calculating Electric Field Intensity Midway Between Two Opposite Charges

Understanding electric field intensity and its calculation is crucial in the realms of physics and engineering. In this article, we will explore how to determine the electric field intensity at a point midway between two charges of opposite signs. We'll follow the principles of superposition and the formula for the electric field due to a point charge.

Introduction to Electric Fields

The electric field, denoted as ( E ), is a vector field that associates to each point in space the force per unit charge that would be exerted on a stationary, point-like electric charge. It is an essential concept in electrostatics, particularly when dealing with charged particles.

Electric Field Due to a Point Charge

The electric field ( E ) due to a point charge ( Q ) at a distance ( r ) from the charge is given by the formula:

[ E frac{kQ}{r^2} ]

where ( k ) is Coulomb's constant, approximately ( 8.99 times 10^9 text{ N m}^2/text{C}^2 ), and ( Q ) is the charge in Coulombs. The direction of the electric field depends on the sign of the charge: the field points away from the positive charge and towards the negative charge.

Example Calculation

Consider two charges, ( Q_1 2.0 times 10^{-7} text{ C} ) and ( Q_2 -2.0 times 10^{-7} text{ C} ), which are separated by a distance of 100 cm (1.0 m). We need to find the electric field intensity at point ( P ), which is midway between the two charges.

Determining the Midpoint Distance

The distance from each charge to point ( P ), the midpoint, can be calculated as:

[ r frac{d}{2} frac{1.0 text{ m}}{2} 0.5 text{ m} ]

Calculating the Electric Field for Each Charge

First, let's calculate the electric field due to ( Q_1 ), the positive charge:

[ E_1 frac{kQ_1}{r^2} frac{8.99 times 10^9 times 2.0 times 10^{-7}}{(0.5)^2} frac{8.99 times 10^9 times 2.0 times 10^{-7}}{0.25} 7192 text{ N/C} ]

The direction of ( E_1 ) is away from the positive charge, directed towards point ( P ).

Next, let's calculate the electric field due to ( Q_2 ), the negative charge:

[ E_2 frac{kQ_2}{r^2} frac{8.99 times 10^9 times 2.0 times 10^{-7}}{(0.5)^2} frac{8.99 times 10^9 times 2.0 times 10^{-7}}{0.25} 7192 text{ N/C} ]

The direction of ( E_2 ) is towards the negative charge, also directed towards point ( P ).

Total Electric Field at Point P

Since both electric fields ( E_1 ) and ( E_2 ) are directed towards point ( P ), they add up:

[ E_{text{total}} E_1 E_2 7192 text{ N/C} 7192 text{ N/C} 14384 text{ N/C} ]

Hence, the electric field intensity at point ( P ) is ( 14384 text{ N/C} ), directed towards the midpoint between the two charges.

Conclusion

By using the principle of superposition and the formula for the electric field due to a point charge, we can determine the electric field intensity at a point midway between two charges of opposite signs. The electric field intensities add up in the scenario where the charges are equal in magnitude but opposite in sign.

Keywords: electric field, electric field intensity, point charge