A 7.0-μC point charge and a
Point charge are initially extremely far apart.How much work does it take to bring the
Point charge to the point

,and the
Point charge to the point

(k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Question

A 7.0-μC point charge and a A 7.0-μC point charge and a Point charge are initially extremely far apart.How much work does it take to bring the Point charge to the point ,and the Point charge to the point (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2) A) 95 J B) 190 J C) 63 J D) 16 J

A 7.0-μC point charge and a Point charge are initially extremely far apart.How much work does it take to bring the Point charge to the point ,and the Point charge to the point (k = 1/4πε<sub>0</sub> = 9.0 × 10<sup>9</sup> N ∙ m<sup>2</sup>/C<sup>2</sup>) A) 95 J B) 190 J C) 63 J D) 16 J

Point charge are initially extremely far apart.How much work does it take to bring the A 7.0-μC point charge and a Point charge are initially extremely far apart.How much work does it take to bring the Point charge to the point ,and the Point charge to the point (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2) A) 95 J B) 190 J C) 63 J D) 16 J

Point charge to the point A 7.0-μC point charge and a Point charge are initially extremely far apart.How much work does it take to bring the Point charge to the point ,and the Point charge to the point (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2) A) 95 J B) 190 J C) 63 J D) 16 J

,and the

Point charge to the point A 7.0-μC point charge and a Point charge are initially extremely far apart.How much work does it take to bring the Point charge to the point ,and the Point charge to the point (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2) A) 95 J B) 190 J C) 63 J D) 16 J

(k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Quizplus · Accepted Answer

To bring a point charge from infinity to a point, the work done is given by the equation W = (kqQ)/r, where k is the Coulomb constant, q and Q are the magnitudes of the charges, and r is the distance between them.

For the first point charge (7.0 μC), the work done to bring it to the point (1.8 m, 0) is:

W1 = (9.0 × 10^9 N ∙ m^2/C^2)(7.0 μC)(1.0/(1.8 m)) = 35 J

For the second point charge (11.0 μC), the work done to bring it to the point (-1.8 m, 0) is:

W2 = (9.0 × 10^9 N ∙ m^2/C^2)(11.0 μC)(1.0/(1.8 m)) = 60 J

Therefore, the total work done to bring both point charges to their respective points is:

Wtot = W1 + W2 = 35 J + 60 J = 95 J

Hence, the answer is A) 95 J.

A 7.0-μC Point Charge and a