show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [hint: for (a), use gauss’s law. for, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
net work done in carrying charge along closed loop by electric field is zero. hence, the tangential component of the electric field is continuous from one side of a charged surface to the other.
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here, when one spring is compressed, the other is stretched and the potential energy will be stored in the compressed spring and so the oscillation continues and then the other spring compresses and the process goes on considering the surface to be a frictionless surface. the oscillation goes on forever if there are no dissipative forces. that depends on a lot of things. in particular, you would need to find the electric susceptibility of the paper. the field produced by static charge is inverse-square. the polarization of the paper varies directly with the external field, with a proportionality constant equal to the electric susceptibility. the dipole field created by the polarized paper is inverse-cube. overall, the force between the ruler and the paper will vary with the inverse fifth power of distance. to get any more detailed, you would need to know the geometry of the situation, and the charge distribution. yes. let us consider a block a which is resting on another block b. block b is resting on a smooth horizontal surface. let the coefficient of friction between the blocks be \[\mu\] . when a force f is applied on block b in the forward direction as shown in the above figure, block a moves with block b in the direction of the applied force. the frictional force on block a and the displacement will be in the forward direction. therefore, work done by the frictional force is positive. if we consider the reference frame of block b, then displacement of block a will be zero. therefore, work done by the frictional force is zero. &nbsp;an electron enters uniform electric field maintained by parallel plates and of value 'e' &nbsp; v m-1 with a velocity ' v ' ms-1.&nbsp;the plates are separated by a distance 'd' metre. what is the acceleration of the electron in the field?&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; dear student,
we know that,after the discovery of electrons, we got to know that
electrons are the particles that flow in a conductor electrons
being the negatively charged particles flow from the negative
terminal to the positive terminal and for current now, electric or
conventional current is the rate of flow of charge so, the
direction of the current should be same as the direction of the
charge, i e positive charge(conventionally), so the direction is
opposite to the flow of the negative charge(flow of electrons)
hence, the direction of the flow of electrons is opposite to the
direction of electric(or conventional) current
regards coulomb's law:f=kq1q2r2, where k is the coulomb's constant and r is the separation between the two charges. the magnitude of the net force:fnet=fx2+fy2k = 9 × 109 n·m2/c2the force due to charge q1 is:f1a=kq1qar1a2=(9×109)(21.4×10-6)(13.5×10-6)(80.1×10-2)2[readmore]f1a = 4.05 nthe force due to charge, q2:f2a=kq2qar2a2=(9×109)(30.6×10-6)(13.5×10-6)[(80.1×10-2)2+(80.1×10-2)2]2f2a = 2.9 nforce due to charge, q3:f3a=kq3qar3a2=(9×109)(57.3×10-6)(13.5×10-6)(80.1×10-2)2f3a = 10.9 nthe net force along the x-axis:fx=f1a-f2acos(45°)=4.05-(2.9)cos(45°)fx = 2.0 nthe net force in the y-direction:fy=f3a-f2acos(45°)=10.9-(2.9)cos(45°)fy = 8.85 nthe magnitude of the net force:fnet=fx2+fy2=2.02+8.852fnet = 9.07 nthe magnitude of the net force on the charge at point a is 9.07 n. The answer to the question is available in the PDF file https://www.assignmentexpert.com/https://www.assignmentexpert.com/homework-answers/physics-answer-3862.pdf from the figure:total resistance of the circuit, r = (rab + 15) ω = 20 ωcurrent in the circuit , \[i = \frac{v}{r} = \frac{8}{20} a\] ∴ voltage across ab, vab = i.rab = 2 vthe emf of the cell connected as above is given by: \[e = \frac{l}{l} v\_0\] here:l = 60 cm (balance point)ab = l = 1 m = 100 cm (total length of the wire) \[\therefore e = \frac{60}{100}\left( 2 \right) = 1 . 2 v\] The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 All that is needed is some sort of wave-shape that sweeps the oscilloscope beam to and fro. Since the relationship between vertical displacement and horizontal displacement on the oscilloscope’s trace is purely a function of the test component’s characteristic curve, we really don’t care what form of [AC](https://www.allaboutcircuits.com/textbook/alternating-current/chpt-1/what-is-alternating-current-ac/) voltage excites it.