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Two tiny conducting spheres are identical and carry charges of -20μC and +50μC. They are separeted by a distance of 2.50cm. (a) what is the magnitude of the force that each sphere each sphere experience, and is the force attractive or repulsive ? (b) The spheres are brought into contact and then separated toa distance of 2.50cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.

User Rob Winch
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1 Answer

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Answer:


14400\ \text{N}, Attractive


3240\ \text{N}, Repulsive

Step-by-step explanation:


q_1 = -20 μC


q_2 = 50 μC

r = Distance between the charges = 2.5 cm

k = Coulomb constant =
9* 10^9\ \text{Nm}^2/\text{C}^2

Electrical force is given by


F=(kq_1q_2)/(r^2)\\\Rightarrow F=(9* 10^9* (-20* 10^(-6))* (50* 10^(-6)))/((2.5*10^(-2))^2)\\\Rightarrow F=-14400\ \text{N}

The magnitude of force each sphere will experience is
14400\ \text{N}

Since the charges have opposite charges they will attract each other.

Now the charges are brought into contact with each other so the resultant charge will be


q=(q_1+q_2)/(2)\\\Rightarrow q=(-20+50)/(2)\\\Rightarrow q=15\ \mu\text{C}


F=(kq^2)/(r^2)\\\Rightarrow F=(9* 10^9* (15* 10^(-6))^2)/((2.5* 10^(-2))^2)\\\Rightarrow F=3240\ \text{N}

The magntude of the force the spheres experience will be
3240\ \text{N}

The spheres have the same charge now so they will repel each other.

User Emanegux
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