Question

Two charged balls are 10 cm apart. One is positively charged, and the other is negatively charged. Describe the electric field lines and any forces that would exist between the two balls.

Answer 1

The electric field lines between two charged balls with opposite charges extend from the positive to the negative. A positive test charge in this field would be repelled by the positive charge and attracted to the negative charge, exhibiting an attractive Coulomb force between the balls.

Step-by-step explanation:

When two charged balls are 10 cm apart, one with a positive charge and the other with a negative charge, the electric field lines and forces between them can be described as follows:

• The electric field lines originate from the positively charged ball and terminate on the negatively charged ball.
• A positive test charge placed within this field would experience a force directing it away from the positive ball and towards the negative ball, consistent with Coulomb's law.
• The force between the two charged balls is attractive, as one ball is positively charged and the other is negatively charged, creating an attractive Coulomb force between them.

The strength of this force can be calculated using Coulomb's law, which states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

Answer 2

The field lines extend from the positive to negative charge.

Force between them is attractive.

Step-by-step explanation:

• The field lines of the electric charges extend from the positive to negative side charge. These are the parallel lines and they never cross each other when tho charges are kept infinitesimal distance apart.
• The distance between the two charges, x = 10 cm
• The force between the two charged balls is attractive.
• It is directly proportional to the product of their charges and inversely proportional to the magnitude of the distance between them.
• It is given by the formula

`F=(1)/(4\pi\epsilon_(0))(q_(1)q_(2))/(x^(2) )` C

• Since distance between the two charges, x = 10 cm, the force between the charges is reduced by the factor 0.01 m² in SI units.