A metallic sphere, with a radius of 10 cm, has an electric field strength of 3600 N/C on its surface. Another metallic sphere, with a radius of a 5 cm, has the same electric field strength on its surface. These two spheres are connected by a long, thin, metal wire. What is the electric field strength on the surface of the larger sphere after two spheres are connected?
Question
A metallic sphere, with a radius of 10 cm, has an electric field strength of 3600 N/C on its surface. Another metallic sphere, with a radius of a 5 cm, has the same electric field strength on its surface. These two spheres are connected by a long, thin, metal wire. What is the electric field strength on the surface of the larger sphere after two spheres are connected?
Solution
The electric field strength on the surface of a metallic sphere is given by E = kQ/R^2, where E is the electric field strength, k is Coulomb's constant (9 x 10^9 Nm^2/C^2), Q is the charge on the sphere, and R is the radius of the sphere.
Given that the electric field strength on the surface of both spheres is the same before they are connected, we can set up the following equation:
E1 = E2 kQ1/R1^2 = kQ2/R2^2
We know that R1 = 10 cm = 0.1 m and R2 = 5 cm = 0.05 m. We can solve for Q1 and Q2:
Q1 = E1 * R1^2 / k Q2 = E2 * R2^2 / k
Since E1 = E2 = 3600 N/C, we can simplify to:
Q1 = 3600 N/C * (0.1 m)^2 / (9 x 10^9 Nm^2/C^2) = 4 x 10^-6 C Q2 = 3600 N/C * (0.05 m)^2 / (9 x 10^9 Nm^2/C^2) = 1 x 10^-6 C
When the two spheres are connected by a wire, the charges on the spheres will redistribute until the electric field strength on the surfaces of both spheres is the same. The total charge Q_total = Q1 + Q2 = 5 x 10^-6 C.
The electric field strength on the surface of the larger sphere after the spheres are connected is then given by:
E_total = kQ_total / R1^2 = (9 x 10^9 Nm^2/C^2) * (5 x 10^-6 C) / (0.1 m)^2 = 4500 N/C
So, the electric field strength on the surface of the larger sphere after the two spheres are connected is 4500 N/C.
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