A -8.70 nC charge of mass 1.30×10-6 kg is released from rest 4.20 mm above a very large plane of positive charge. The charge accelerates toward the plane and collides with a speed of 3.20 m/s. What is the surface charge density of the plane? 1.01×10-6 C/m2 2.25×10-7 C/m2 1.61×10-6 C/m2 3.22×10-6 C/m2
Question
A -8.70 nC charge of mass 1.30×10-6 kg is released from rest 4.20 mm above a very large plane of positive charge. The charge accelerates toward the plane and collides with a speed of 3.20 m/s. What is the surface charge density of the plane? 1.01×10-6 C/m2 2.25×10-7 C/m2 1.61×10-6 C/m2 3.22×10-6 C/m2
Solution
To solve this problem, we need to use the principle of conservation of energy. The initial potential energy of the charge is converted into kinetic energy when it hits the plane.
The potential energy (PE) of a charge in an electric field is given by PE = qV, where q is the charge and V is the potential (voltage). The electric potential of an infinite plane of charge is given by V = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space (8.85 x 10^-12 C^2/Nm^2).
The kinetic energy (KE) of an object is given by KE = 1/2mv^2, where m is the mass and v is the speed.
Setting the initial potential energy equal to the final kinetic energy gives us:
qσ/2ε0 = 1/2m*v^2
Solving for σ gives us:
σ = 2mv^2*ε0/q
Substituting the given values:
σ = 2*(1.30 x 10^-6 kg)(3.20 m/s)^2(8.85 x 10^-12 C^2/N*m^2)/(-8.70 x 10^-9 C)
Solving this gives us a surface charge density of approximately 1.01 x 10^-6 C/m^2. So, the correct answer is 1.01 x 10^-6 C/m^2.
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