A pendulum of mass \(80mg\) and carrying a charge of \(20nC\) is at rest in a horizontal uniform electric field of \(\left(2\times10^{4}\right)\frac{V}{m}\). Find: the tension in the tread of pendulum and angle it makes with the vertical.
Ans: \(\theta=tan^{-1}\left(0.5102\right)=27^{o}\) and \(T=\left(8.81\times10^{-4}\right)N\).
An electron is being liberated from the lower of two large parallel metal plates separated by a distance of \(0.02m\). The upper plate has a potential of \(\left(2.4\times10^{3}\right)V\) relative to the lower plate. How long does the electron take to reach the upper plate? Take: \(\frac{e}{m}\) of electron as An electron is being liberated from the lower of two large parallel metal plates separated by a distance of \(0.02m\). The upper plate has a potential of \(\left(2.4\times10^{3}\right)V\) relative to the lower plate. How long does the electron take to reach the upper plate? Take: \(\frac{e}{m}\) of electrons as \(\left(1.8\times10^{11}\right)\frac{C}{kg}\).
Ans: \(t=\left(1.4\times10^{-9}\right)Sec\).
A particle of mass \(m\) and charge \(q\) is released from rest in a uniform electric field of intensity \(E\).Calculate: the kinetic energy it attains after moving a distance \(X\) between the plates.
Ans: \(K.E=EqX\).
A particle of mass m kg and charge q C is thrown at a speed of V \(\frac{m}{s}\) against a uniform electric field of strength E \(\frac{N}{C}\). How much distance will be travelled by the particle before coming to rest?
Ans: \(S=\frac{v^{2}.m}{2q.E}\).
An electron falls through a distance of \(1.5cm\) in a uniform electric field of magnitude of \(\left(2.0\times10^{4}\right)\frac{N}{C}\) as shown in figure. The direction of the field is reversed keeping its magnitude same as of previous and a proton falls through the same distance. Compute: the time of fall in each case. Contrast the situation first with that of the “free fall under gravity”.
Ans: \(t_{electron}=\left(2.9\times10^{-9}\right)Sec\) and \(t_{proton}=\left(1.25\times10^{-7}\right)Sec\).
An electric field E is st up between two parallel plates of a capacitor as shown in given figure. An electron enters the field symmetrically between the plates with a speed \(v_{o}\). The length of the each side is \(L\). Find the angle of deviation of the path of the electron as it comes out of the field.
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of \(\left(25.5\times10^{3}\right)\frac{V}{m}\). The density of the liquid is \(\left(1.26\times10^{3}\right)\frac{kg}{m^{3}}\). Find: the radius of the drop.
Ans: \(r=\left(4.3\times10^{-7}\right)m\).
A wall which is uniformly charged and providing a uniform electric field normally of \(\left(4\times10^{9}\right)\frac{N}{C}\). A charged particle of mass \(2g\) is suspended through a silk thread of length \(24cm\) and of negligible mass, stays at a distance of \(12cm\) from wall. Find: the charge on particle.[Given:\(g=10\frac{m}{s^{2}}\)].
Ans: \(q=0.289\mu C\).
Two point charges \(+2\mu C\) and \(-2\mu C\) are located \(10cm\) apart from each other in air. (a) Calculate: the electric field at the mid point P of the line joining the two charges. (b) If, a negative charge of \(\left(1.6\times10^{-9}\right)C\) is placed at that point. Find: the force experienced by this charge.
Ans:(a) \(E=\left(14.4\times10^{6}\right)\frac{N}{C}\) and (b) \(F=-\left(2.3\times10^{-12}\right)N\).
Two point charges \(+4\mu C\) and \(+1\mu C\) are separated by a distance of \(2m\) in air. Find: the point on the line joining charges at which the net electric field of the system is Zero.
Ans: \(\frac{4}{3}m\)
Two point charges of \(\left(16\times10^{-6}\right)C\) and \(-\left(9\times10^{-6}\right)C\) are placed \(8cm\) apart in air. Determine: the position of the point at which the resultant field is Zero.
Ans: \(24cm \) right to the charge \(-\left(9\times10^{-6}\right)C\).
Four charges \(+q\), \(+q\), \(-q\) and \(-q\) are placed respectively at the four corner of a square of side \(a\). Find: the magnitude and direction of the electric field at the center of the square.
Two point charges \(+q\) and \(-2q\) are placed at the vertices B and C of an equilateral triangle ABC of side a. Obtain: expression for magnitude and direction of resultant electric field at the vertex A due to these two charges.
Ans: \(E_{R}=E\times\sqrt{3}=\frac{1}{4\pi\epsilon_{o}}\frac{q}{a^{2}}\sqrt{3}\) and \(\beta=tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{o}\).
Two point charges \(q_{1}\) and \(q_{2}\) of \(10^{-8}C\) and \(-10^{-8}C\) respectively are placed \(0.1m\) apart. Calculate: the electric field at points A, B and C as shown in the figure.
Ans: \(E_{A}=\left(7.2\times10^{4}\right)\frac{N}{C}\), \(E_{B}=\left(3.2\times10^{4}\right)\frac{N}{C}\) and \(E_{C}=\left(9\times10^{3}\right)\frac{N}{C}\).
PQRS is a square having side \(5m\). If charges \(+50C\), \(-50C\) and \(+50C\) are placed at P, R and S respectively. What is the resultant electric field at Q?
Ans: \(E=\left(2.7\times10^{10}\right)\frac{N}{C}\) and \(\beta=25.5^{o}\).
