What is the the electric field at a point midway between a -7.67 μc and a +5.03 μc charge 3.38 cm apart? take the direction towards the positive charge to be positive.

-4.01 × 10⁸ N/C.

Explanation:

Given : Charge Q = -7.67 × 10⁻⁶ C

Charge q = + 5.03×10⁻⁶ C

Distance between the charges = 0.0338 m

Let d be the distance from the charge Q and charge q to the mid point.

Midway distance =  d = 0.0169 m

Electric field due to the negative charge = E₁ = k Q / d²

Electric field due to the positive charge = E₂ = k q / d²

Here k = 9 × 10⁹ N m²/C² is the Coulomb's constant.

E₁ = (9 × 10⁹)(7.67 × 10⁻⁶) / (0.0169)² = 2.42 × 10⁸ N/C

E₂ = (9 × 10⁹)( 5.03×10⁻⁶) /(0.0169)² = 1.585 × 10⁸N/C

Assume that the negative charge is  towards the left of the mid point and the positive charge is towards the right of the mid point,

The electric field due to the negative charge is inwards towards the charge(left) and the electric field due to positive charge is outwards and so towards the negative X direction.

Total electric field = E = E₁ + E₂ = -2.42 × 10⁸ - 1.585 × 10⁸

                                                     = - 4.01 × 10⁸ N/C

Direction towards the left (towards the negative charge).

heat the gas and measure the pressure