A \(m=20\,\mathrm{g}\) mass with charge \(q=+8\,\mathrm{\mu C}\) hangs from a spring of stiffness \(k=50\,\mathrm{N/m}\). If an infinite plane with charge density \(\sigma=+3\,\mathrm{\mu C/m^2}\) is placed \(d=5\,\mathrm{cm}\) below the masses original position, through what distance does the mass on the spring deflect in \(\,\mathrm{m}\)? Ignore the sign of the deflection and answer with a positive number.
To find the deflection of the mass on the spring due to the electric field created by the infinite plane, we can use the concept of electric potential energy and its relationship with the spring potential energy.
Here are the steps to follow:
Step 1: Find the electric potential energy of the charged mass:
The electric potential energy (\(PE_e\)) of a charged mass is given by the equation:
\[ PE_e = \frac{1}{2} k x^2 \]
where \(k\) is the spring stiffness and \(x\) is the deflection of the mass on the spring.
Step 2: Find the electric potential energy in terms of the electric field:
The electric field (\(E\)) created by an infinite plane is given by the equation:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \(\sigma\) is the charge density of the plane and \(\epsilon_0\) is the permittivity of free space.
The electric potential energy due to the electric field is given by the equation:
\[ PE_e = q Ed \]
where \(q\) is the charge of the mass and \(d\) is the distance between the plane and the original position of the mass.
Step 3: Set the two expressions for the electric potential energy equal to each other and solve for \(x\):
\[\frac{1}{2} k x^2 = q Ed\]
Step 4: Substitute the given values into the equation and solve for \(x\):
Given:
\(m = 20\, \mathrm{g} = 0.02\, \mathrm{kg}\)
\(q = +8\, \mu\mathrm{C} = 8 \times 10^{-6}\, \mathrm{C}\)
\(k = 50\, \mathrm{N/m}\)
\(\sigma = +3\, \mu \mathrm{C/m^2} = 3\times 10^{-6}\, \mathrm{C/m^2}\)
\(d = 5\, \mathrm{cm} = 0.05\, \mathrm{m}\)
Substituting the values:
\[\frac{1}{2} \times 50 \times x^2 = 8 \times 10^{-6} \times 3\times 10^{-6} \times 0.05\]
Simplifying the equation yields:
\[25x^2 = 1.2\times 10^{-10}\]
Step 5: Solve for \(x\):
\[x = \sqrt{\frac{1.2 \times 10^{-10}}{25}}\]
Calculating the value gives:
\[x \approx 2.19 \times 10^{-6}\, \mathrm{m}\]
Therefore, the mass on the spring deflects approximately \(2.19 \times 10^{-6}\, \mathrm{m}\) in response to the presence of the charged plane.