A+7.5 nc point charge and a -2.0 nc point charge are 3.0 cm apart. what is the electric field strength at the midpoint between the two charges?

Ep= 3.8 10⁵ N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ =+7.5 nC = +7.5*10⁻⁹C  

q₂ =  -2.0 nC = -2.0*10⁻⁹C

d₁ =d₂ = 1.5cm = 1.5 *10⁻²m  = 0.015 m

Calculation of the electric fieldsat the midpoint (P) between the two charges

Look at the attached graphic:

E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .

E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge

E₁ = k*q₁/d₁² = 9*10⁹ *7.5  *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep= E₁ + E₂  

Ep= 3*10⁵ N/C +  0.8*10⁵ N/C

Ep= 3.8 10⁵ N/C

b. false

explanation:

suppose a rock in free fall is dropped off a cliff. after falling for 5 seconds, its speed would not be 100 m/s.