Two point charges are placed 3 m apart. One charge is -1.5 nC and the other is +2 nC. Calculate the electric field strength at the midpoint between the two charges.

Steps to Solve

1. Identify the midpoint and distances The two charges are separated by 3 m, so the midpoint is at 1.5 m from each charge.

2. Use the formula for electric field strength The electric field strength $E$ due to a point charge is given by the formula: $$ E = \frac{k \cdot |Q|}{r^2} $$ where:

• ( k ) is Coulomb's constant, approximately ( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 ),
• ( Q ) is the charge,
• ( r ) is the distance from the charge to the point of interest.

3. Calculate the electric field due to each charge at the midpoint For the negative charge ( -1.5 , \text{nC} ):

• Convert the charge to Coulombs: ( Q_1 = -1.5 \times 10^{-9} , \text{C} ).
• Distance to midpoint ( r_1 = 1.5 , \text{m} ).
• Electric field due to ( Q_1 ): $$ E_1 = \frac{8.99 \times 10^9 \cdot | -1.5 \times 10^{-9}|}{(1.5)^2} $$

For the positive charge ( +2 , \text{nC} ):

• Convert the charge to Coulombs: ( Q_2 = 2 \times 10^{-9} , \text{C} ).
• Distance to midpoint ( r_2 = 1.5 , \text{m} ).
• Electric field due to ( Q_2 ): $$ E_2 = \frac{8.99 \times 10^9 \cdot | 2 \times 10^{-9}|}{(1.5)^2} $$

4. Calculate individual electric fields For ( E_1 ): $$ E_1 = \frac{8.99 \times 10^9 \cdot 1.5 \times 10^{-9}}{2.25} = 5.99 \times 10^0 , \text{N/C} $$ (direction toward the charge)

For ( E_2 ): $$ E_2 = \frac{8.99 \times 10^9 \cdot 2 \times 10^{-9}}{2.25} = 7.99 \times 10^0 , \text{N/C} $$ (direction away from the charge)

5. Determine the resultant electric field Since ( E_1 ) is directed toward the negative charge and ( E_2 ) is directed away from the positive charge, we sum their magnitudes because they are in opposite directions: $$ E_{\text{net}} = E_2 - E_1 = 7.99 - 5.99 = 2.00 , \text{N/C} $$