An electron is acted on by two electric forces, one of 2.7×10-14 N acting upward and a second of 5.8×10-14 N acting to the right. What is the magnitude of the electric field at the electron’s location?

3.99×10⁵ N/C

Explanation:

Assuming both force forms a right angle triangle,

Using Pythagoras theorem,

a² = b²+c² Equation 1

Where a= resultant of the two forces, b = the force acting upward, c = the force acting to the right.

make a the subject of the equation.

a = √(b²+c²) Equation 2

Given: b = 2.7×10⁻¹⁴ N, c = 5.8×10⁻¹⁴ N.

Substitute into equation 2

a = √[(2.7×10⁻¹⁴)²+( 5.8×10⁻¹⁴ )²]

a = √[7.28×10⁻²⁸)+(33.64×10⁻²⁸)]

a = √(40.92×10⁻²⁸)

a = 6.397×10⁻¹⁴ N.

But,

F = Eq Equation 3

Where E = Electric Field, q = charge of an electron, F = electric force.

make E the subject of the equation,

E = F/q Equation 4

Given: F = 6.397×10⁻¹⁴ N, q = 1.602×10⁻¹⁹  C.

Substitute into equation 4

E =  6.397×10⁻¹⁴/(1.602×10⁻¹⁹ )

E = 3.99×10⁵ N/C

reserve is the answer