Two tiny particles having charges +20.0 ?C and -8.00 ?C are separated by a distance of 20.0 cm. What are the magnitude and direction of electric field midway between these two charges? (k = 1/4??₀ = 9.0 × 10⁹ N ? m²/C²) A)25.2 × $10 ^ { 6 }$ N/C directed towards the negative charge B)25.2 × $10 ^ { 6 }$ N/C directed towards the positive charge C)25.2 × $10 ^ { 5 }$ N/C directed towards the negative charge D)25.2 × $10 ^ { 5 }$ N/C directed towards the positive charge E)25.2 × $10 ^ { 4 }$ N/C directed towards the negative charge

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The electric field due to a point charge is given by:$$E = k\frac{q}{r^2}$$where:* k is Coulomb's constant (9.0 × 10^9 N m^2/C^2)* q is the charge of the point charge (in Coulombs)* r is the distance from the point charge (in meters)In this problem, we have two point charges, one with a charge of +20.0 μC and the other with a charge of -8.00 μC. The distance between the two charges is 20.0 cm. We want to find the electric field midway between the two charges.The electric field due to the positive charge is:$$E_1 = k\frac{q_1}{r^2} = (9.0 	imes 10^9 	ext{ N m}^2/	ext{C}^2)\frac{20.0 	imes 10^{-6} 	ext{ C}}{(0.200 	ext{ m})^2} = 4.50 	imes 10^6 	ext{ N/C}$$The electric field due to the negative charge is:$$E_2 = k\frac{q_2}{r^2} = (9.0 	imes 10^9 	ext{ N m}^2/	ext{C}^2)\frac{-8.00 	imes 10^{-6} 	ext{ C}}{(0.200 	ext{ m})^2} = -1.80 	imes 10^6 	ext{ N/C}$$The net electric field is the vector sum of the electric fields due to the two charges:$$E = E_1 + E_2 = 4.50 	imes 10^6 	ext{ N/C} - 1.80 	imes 10^6 	ext{ N/C} = 2.70 	imes 10^6 	ext{ N/C}$$The direction of the net electric field is towards the negative charge.Therefore, the correct answer is A: 25.2 × 10^6 N/C directed towards the negative charge.

Two Tiny Particles Having Charges +20 10610 ^ { 6 }106

Two Tiny Particles Having Charges +20 $10 ^ { 6 }$