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Question 1
Two small metallic spheres are brought into contact. Sphere A initially carries a charge of $+ 8 μ C$ and sphere B carries a charge of $- 4 μ C$ . After contact, what charge does each sphere carry?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $+ 2 μ C$ each
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the initial charges
Sphere A carries $+ 8 μ C$ and Sphere B carries $- 4 μ C$ . When identical metallic spheres touch, charge redistributes equally between them.
Step 2: Apply conservation of charge
Total charge before contact: $+ 8 + (- 4) = + 4 μ C$ . By conservation of charge, the total charge after contact must also be $+ 4 μ C$ .
Step 3: Distribute charge equally
Since the spheres are identical, charge distributes equally: each sphere receives $\frac{+ 4}{2} = + 2 μ C$ .
Step 4: State the result
Each sphere ends up with $+ 2 μ C$ . The total charge is conserved: $2 + 2 = + 4 μ C$ , consistent with the initial total.
Method #2Approach 2
Step 1: Identify what is being asked
We need the final charge on each identical sphere after contact. Two key principles apply: conservation of charge and equal sharing between identical conductors.
Step 2: Eliminate '$+4\,\mu\mathrm{C}$ each'
This gives a total of $+ 8 μ C$ , but the initial total is only $+ 4 μ C$ . This violates conservation of charge.
Step 3: Eliminate '$-2\,\mu\mathrm{C}$ each'
This gives a total of $- 4 μ C$ , but the initial total is $+ 4 μ C$ . Again, charge is not conserved.
Step 4: Eliminate '$+6\,\mu\mathrm{C}$ and $-2\,\mu\mathrm{C}$'
Although the total is $+ 4 μ C$ (charge conserved), identical spheres share charge equally — they cannot retain different charges after contact.
Step 5: Select '$+2\,\mu\mathrm{C}$ each'
Total charge $+ 4 μ C$ shared equally gives $+ 2 μ C$ each, satisfying both conservation and equal sharing.
Question 2
Two point charges $q_{1} = + 4 μ C$ and $q_{2} = + 9 μ C$ are separated by a distance of $0.06 m$ . What is the magnitude of the electrostatic force between them?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $90 N$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the given values
$q_{1} = 4 \times 1 0^{- 6} C$ , $q_{2} = 9 \times 1 0^{- 6} C$ , $r = 0.06 m$ , $k = 8.99 \times 1 0^{9} N m^{2} C^{- 2}$ .
Step 2: Apply Coulomb's law
$F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 4 \times 1 0 ^{- 6} ) ( 9 \times 1 0 ^{- 6} )}{( 0.06 ) ^{2}}$
Step 3: Calculate numerator and denominator
Numerator: $(8.99 \times 1 0^{9}) (36 \times 1 0^{- 12}) = 8.99 \times 36 \times 1 0^{- 3} = 323.64 \times 1 0^{- 3} = 0.3236 N m^{2}$ . Denominator: $(0.06)^{2} = 3.6 \times 1 0^{- 3} m^{2}$ .
Step 4: Compute the result
$F = \frac{0.3236}{3.6 \times 1 0 ^{- 3}} \approx 89.9 N \approx 90 N$ . Both charges are positive, so the force is repulsive.
Method #2Approach 2
Step 1: Identify the approach
We apply $F = k ∣ q_{1} q_{2} ∣/ r^{2}$ . The key step is correctly squaring $r = 0.06 m$ to get $r^{2} = 3.6 \times 1 0^{- 3} m^{2}$ .
Step 2: Eliminate '$0.90\,\mathrm{N}$'
This is 100 times too small. It could result from forgetting to convert $μ C$ to C properly or making an order-of-magnitude error.
Step 3: Eliminate '$9.0\,\mathrm{N}$'
This is 10 times too small, likely arising from an error in powers of 10 during the calculation.
Step 4: Eliminate '$45\,\mathrm{N}$'
This is approximately half the correct answer, which might result from incorrectly using $r$ instead of $r^{2}$ in the denominator or halving the product of charges.
Step 5: Select '$90\,\mathrm{N}$'
The correct computation gives $F \approx 90 N$ . This is the repulsive force between the two positive charges.

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13 more questions in this topic

Free preview 2/15

Question 1
Two small metallic spheres are brought into contact. Sphere A initially carries a charge of $+ 8 μ C$ and sphere B carries a charge of $- 4 μ C$ . After contact, what charge does each sphere carry?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $+ 2 μ C$ each
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the initial charges
Sphere A carries $+ 8 μ C$ and Sphere B carries $- 4 μ C$ . When identical metallic spheres touch, charge redistributes equally between them.
Step 2: Apply conservation of charge
Total charge before contact: $+ 8 + (- 4) = + 4 μ C$ . By conservation of charge, the total charge after contact must also be $+ 4 μ C$ .
Step 3: Distribute charge equally
Since the spheres are identical, charge distributes equally: each sphere receives $\frac{+ 4}{2} = + 2 μ C$ .
Step 4: State the result
Each sphere ends up with $+ 2 μ C$ . The total charge is conserved: $2 + 2 = + 4 μ C$ , consistent with the initial total.
Method #2Approach 2
Step 1: Identify what is being asked
We need the final charge on each identical sphere after contact. Two key principles apply: conservation of charge and equal sharing between identical conductors.
Step 2: Eliminate '$+4\,\mu\mathrm{C}$ each'
This gives a total of $+ 8 μ C$ , but the initial total is only $+ 4 μ C$ . This violates conservation of charge.
Step 3: Eliminate '$-2\,\mu\mathrm{C}$ each'
This gives a total of $- 4 μ C$ , but the initial total is $+ 4 μ C$ . Again, charge is not conserved.
Step 4: Eliminate '$+6\,\mu\mathrm{C}$ and $-2\,\mu\mathrm{C}$'
Although the total is $+ 4 μ C$ (charge conserved), identical spheres share charge equally — they cannot retain different charges after contact.
Step 5: Select '$+2\,\mu\mathrm{C}$ each'
Total charge $+ 4 μ C$ shared equally gives $+ 2 μ C$ each, satisfying both conservation and equal sharing.
Question 2
Two point charges $q_{1} = + 4 μ C$ and $q_{2} = + 9 μ C$ are separated by a distance of $0.06 m$ . What is the magnitude of the electrostatic force between them?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $90 N$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the given values
$q_{1} = 4 \times 1 0^{- 6} C$ , $q_{2} = 9 \times 1 0^{- 6} C$ , $r = 0.06 m$ , $k = 8.99 \times 1 0^{9} N m^{2} C^{- 2}$ .
Step 2: Apply Coulomb's law
$F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 4 \times 1 0 ^{- 6} ) ( 9 \times 1 0 ^{- 6} )}{( 0.06 ) ^{2}}$
Step 3: Calculate numerator and denominator
Numerator: $(8.99 \times 1 0^{9}) (36 \times 1 0^{- 12}) = 8.99 \times 36 \times 1 0^{- 3} = 323.64 \times 1 0^{- 3} = 0.3236 N m^{2}$ . Denominator: $(0.06)^{2} = 3.6 \times 1 0^{- 3} m^{2}$ .
Step 4: Compute the result
$F = \frac{0.3236}{3.6 \times 1 0 ^{- 3}} \approx 89.9 N \approx 90 N$ . Both charges are positive, so the force is repulsive.
Method #2Approach 2
Step 1: Identify the approach
We apply $F = k ∣ q_{1} q_{2} ∣/ r^{2}$ . The key step is correctly squaring $r = 0.06 m$ to get $r^{2} = 3.6 \times 1 0^{- 3} m^{2}$ .
Step 2: Eliminate '$0.90\,\mathrm{N}$'
This is 100 times too small. It could result from forgetting to convert $μ C$ to C properly or making an order-of-magnitude error.
Step 3: Eliminate '$9.0\,\mathrm{N}$'
This is 10 times too small, likely arising from an error in powers of 10 during the calculation.
Step 4: Eliminate '$45\,\mathrm{N}$'
This is approximately half the correct answer, which might result from incorrectly using $r$ instead of $r^{2}$ in the denominator or halving the product of charges.
Step 5: Select '$90\,\mathrm{N}$'
The correct computation gives $F \approx 90 N$ . This is the repulsive force between the two positive charges.

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Step-by-step walkthrough

Step 1: Identify the initial charges

Step 2: Apply conservation of charge

Step 3: Distribute charge equally

Step 4: State the result

Step 1: Identify what is being asked

Step 2: Eliminate '$+4\,\mu\mathrm{C}$ each'

Step 3: Eliminate '$-2\,\mu\mathrm{C}$ each'

Step 4: Eliminate '$+6\,\mu\mathrm{C}$ and $-2\,\mu\mathrm{C}$'

Step 5: Select '$+2\,\mu\mathrm{C}$ each'

Step-by-step walkthrough

Step 1: Identify the given values

Step 2: Apply Coulomb's law

Step 3: Calculate numerator and denominator

Step 4: Compute the result

Step 1: Identify the approach

Step 2: Eliminate '$0.90\,\mathrm{N}$'

Step 3: Eliminate '$9.0\,\mathrm{N}$'

Step 4: Eliminate '$45\,\mathrm{N}$'

Step 5: Select '$90\,\mathrm{N}$'

13 more questions in this topic

Step-by-step walkthrough

Step 1: Identify the initial charges

Step 2: Apply conservation of charge

Step 3: Distribute charge equally

Step 4: State the result

Step 1: Identify what is being asked

Step 2: Eliminate '$+4\,\mu\mathrm{C}$ each'

Step 3: Eliminate '$-2\,\mu\mathrm{C}$ each'

Step 4: Eliminate '$+6\,\mu\mathrm{C}$ and $-2\,\mu\mathrm{C}$'

Step 5: Select '$+2\,\mu\mathrm{C}$ each'

Step-by-step walkthrough

Step 1: Identify the given values

Step 2: Apply Coulomb's law

Step 3: Calculate numerator and denominator

Step 4: Compute the result

Step 1: Identify the approach

Step 2: Eliminate '$0.90\,\mathrm{N}$'

Step 3: Eliminate '$9.0\,\mathrm{N}$'

Step 4: Eliminate '$45\,\mathrm{N}$'

Step 5: Select '$90\,\mathrm{N}$'

13 more questions in this topic