Question

particles of charge +65, +48, and −95 μc are placed in a line (figure 1). the center one is l = 60 cm from each of the others. figure 1 of 1: +65 μc, +48 μc, −95 μc with distance l between each. part a: calculate the net force on the left charge due to the other two. enter a positive value if the force is directed to the right and a negative value if the force is directed to the left. express your answer to two significant figures and include the appropriate units. f_{+65} = value units. submit, request answer. part b: calculate the net force on the center charge due to the other two. enter a positive value if the force is directed to the right and a negative value if the force is directed to the left. express your answer to two significant figures and include the appropriate units.

Explanation:

Part A

Step1: Recall Coulomb's Law

Coulomb's Law is \( F = k\frac{|q_1q_2|}{r^2} \), where \( k = 8.988\times10^{9}\, \text{N·m}^2/\text{C}^2 \), \( q_1, q_2 \) are charges, and \( r \) is the distance between them. The left charge is \( q_1 = +65\,\mu\text{C}=65\times10^{-6}\,\text{C} \), center charge \( q_2 = +48\,\mu\text{C}=48\times10^{-6}\,\text{C} \), right charge \( q_3 = -95\,\mu\text{C}=-95\times10^{-6}\,\text{C} \). Distance from left to center is \( r_1 = L = 0.60\,\text{m} \), from left to right is \( r_2 = 2L = 1.20\,\text{m} \).

Step2: Force from center on left

The center charge (\( +48\,\mu\text{C} \)) repels the left charge (\( +65\,\mu\text{C} \)), so force \( F_{21} \) is to the left (negative direction). Using Coulomb's Law:
\( F_{21} = -k\frac{|q_1q_2|}{r_1^2} = -8.988\times10^{9}\frac{(65\times10^{-6})(48\times10^{-6})}{(0.60)^2} \)
Calculate numerator: \( (65\times10^{-6})(48\times10^{-6}) = 3.12\times10^{-9} \)
Denominator: \( 0.36 \)
\( F_{21} = -8.988\times10^{9}\frac{3.12\times10^{-9}}{0.36} \approx -8.988\times8.6667 \approx -78.0\,\text{N} \)

Step3: Force from right on left

The right charge (\( -95\,\mu\text{C} \)) attracts the left charge (\( +65\,\mu\text{C} \)), so force \( F_{31} \) is to the right (positive direction). Using Coulomb's Law:
\( F_{31} = k\frac{|q_1q_3|}{r_2^2} = 8.988\times10^{9}\frac{(65\times10^{-6})(95\times10^{-6})}{(1.20)^2} \)
Numerator: \( (65\times10^{-6})(95\times10^{-6}) = 6.175\times10^{-9} \)
Denominator: \( 1.44 \)
\( F_{31} = 8.988\times10^{9}\frac{6.175\times10^{-9}}{1.44} \approx 8.988\times4.2882 \approx 38.5\,\text{N} \)

Step4: Net force on left charge

Net force \( F_{\text{net}} = F_{21} + F_{31} = -78.0 + 38.5 = -39.5\,\text{N} \). Rounding to two significant figures: \( -40\,\text{N} \) (or \( -39\,\text{N} \) depending on calculation precision; let's recalculate more accurately).

Recalculating \( F_{21} \):
\( k\frac{q_1q_2}{r_1^2} = 8.988\times10^9 \times \frac{65\times10^{-6} \times 48\times10^{-6}}{0.6^2} = 8.988\times10^9 \times \frac{3.12\times10^{-9}}{0.36} = 8.988\times8.6667 \approx 77.8\,\text{N} \) (repulsive, left: \( -77.8\,\text{N} \))

\( F_{31} = 8.988\times10^9 \times \frac{65\times10^{-6} \times 95\times10^{-6}}{1.2^2} = 8.988\times10^9 \times \frac{6.175\times10^{-9}}{1.44} = 8.988\times4.2882 \approx 38.5\,\text{N} \) (attractive, right: \( +38.5\,\text{N} \))

Net: \( -77.8 + 38.5 = -39.3\,\text{N} \approx -39\,\text{N} \) (two significant figures: \( -39\,\text{N} \) or \( -4.0\times10^1\,\text{N} \))

Part B

Step1: Forces on center charge

Center charge \( q_2 = +48\,\mu\text{C} \). Left charge \( q_1 = +65\,\mu\text{C} \) repels it (force \( F_{12} \) to the right, positive). Right charge \( q_3 = -95\,\mu\text{C} \) attracts it (force \( F_{32} \) to the right, positive, since opposite charges attract: center is +, right is -, so force on center is to the right).

Step2: Force from left on center

\( F_{12} = k\frac{|q_1q_2|}{r_1^2} = 8.988\times10^9\frac{(65\times10^{-6})(48\times10^{-6})}{(0.60)^2} \approx 77.8\,\text{N} \) (right, positive)

Step3: Force from right on center

\( F_{32} = k\frac{|q_2q_3|}{r_2^2} = 8.988\times10^9\frac{(48\times10^{-6})(95\times10^{-6})}{(0.60)^2} \) (distance from center to right is \( L = 0.60\,\text{m} \))
Numerator: \( (48\times10^{-6})(95\times10^{-6}) = 4.56\times10^{-9} \)
Denominator: \( 0.36 \)
\( F_{32} = 8.988\times10^9\frac{4.56\times10^{-9}}{0.36} \approx 8.988\times12.6667 \approx 113.9\,\text{N} \) (right, positive)

Step4: Net force on center charge

Net force \( F_{\text{net}} = F_{12} + F_{32} = 77.8 + 113.9 = 191.7\,\text{N} \). Rounding to two significant figures: \( 1.9\times10^2\,\text{N} \) or \( 190\,\text{N} \) (more accurately, recalculate \( F_{32} \)):

\( F_{32} = 8.988\times10^9 \times \frac{48\times10^{-6} \times 95\times10^{-6}}{0.6^2} = 8.988\times10^9 \times \frac{4.56\times10^{-9}}{0.36} = 8.988\times12.6667 \approx 113.9\,\text{N} \)
\( F_{12} \approx 77.8\,\text{N} \)
Net: \( 77.8 + 113.9 = 191.7 \approx 1.9\times10^2\,\text{N} \) (two significant figures: \( 190\,\text{N} \) or \( 1.9\times10^2\,\text{N} \))

Part A Answer:

\( \boldsymbol{-39\,\text{N}} \) (or \( -4.0\times10^1\,\text{N} \))

Part B Answer:

\( \boldsymbol{1.9\times10^2\,\text{N}} \) (or \( 190\,\text{N} \))

Answer:

Step1: Forces on center charge

Center charge \( q_2 = +48\,\mu\text{C} \). Left charge \( q_1 = +65\,\mu\text{C} \) repels it (force \( F_{12} \) to the right, positive). Right charge \( q_3 = -95\,\mu\text{C} \) attracts it (force \( F_{32} \) to the right, positive, since opposite charges attract: center is +, right is -, so force on center is to the right).

Step2: Force from left on center

\( F_{12} = k\frac{|q_1q_2|}{r_1^2} = 8.988\times10^9\frac{(65\times10^{-6})(48\times10^{-6})}{(0.60)^2} \approx 77.8\,\text{N} \) (right, positive)

Step3: Force from right on center

\( F_{32} = k\frac{|q_2q_3|}{r_2^2} = 8.988\times10^9\frac{(48\times10^{-6})(95\times10^{-6})}{(0.60)^2} \) (distance from center to right is \( L = 0.60\,\text{m} \))
Numerator: \( (48\times10^{-6})(95\times10^{-6}) = 4.56\times10^{-9} \)
Denominator: \( 0.36 \)
\( F_{32} = 8.988\times10^9\frac{4.56\times10^{-9}}{0.36} \approx 8.988\times12.6667 \approx 113.9\,\text{N} \) (right, positive)

Step4: Net force on center charge

Net force \( F_{\text{net}} = F_{12} + F_{32} = 77.8 + 113.9 = 191.7\,\text{N} \). Rounding to two significant figures: \( 1.9\times10^2\,\text{N} \) or \( 190\,\text{N} \) (more accurately, recalculate \( F_{32} \)):

\( F_{32} = 8.988\times10^9 \times \frac{48\times10^{-6} \times 95\times10^{-6}}{0.6^2} = 8.988\times10^9 \times \frac{4.56\times10^{-9}}{0.36} = 8.988\times12.6667 \approx 113.9\,\text{N} \)
\( F_{12} \approx 77.8\,\text{N} \)
Net: \( 77.8 + 113.9 = 191.7 \approx 1.9\times10^2\,\text{N} \) (two significant figures: \( 190\,\text{N} \) or \( 1.9\times10^2\,\text{N} \))

Part A Answer:

\( \boldsymbol{-39\,\text{N}} \) (or \( -4.0\times10^1\,\text{N} \))

Part B Answer:

\( \boldsymbol{1.9\times10^2\,\text{N}} \) (or \( 190\,\text{N} \))