Question

solve question 3,4,5

Answer 1

Q3. Identical cells of EMF E each are connected in Parallel. Internal resistance is negligible. So the positive terminals of both are connected together with a resistance R in between. The negative terminals are connected together.

  Current through the resistance = potential difference across R / resistance
                              I = (E - E)/R = 0.

Q4.  Two charges of magnitude -3Q and +2Q are located at (a,0) and (4a,0)... We need to find the Electric flux Ф_E through a spherical surface of radius R = 5a with center at the origin.

     Answer = q/ε = (-3Q + 2Q)/ε = - Q/ε 

      Apply Gauss's law:  Φ_E = ∫ E · dS = q/ε
          where E = Electric field on the spherical surface
                    dS = Area vector on the spherical surface
                     q = Net Electric Charge enclosed. 
                      ε = charge enclosed.
                      · = dot product of vectors E and dS.
        
     Minus sign for Electric flux means that the electric field lines are flowing inwards from outside. 

5.  Resistance R = dV/di =  1/ (di/dV)
            V = Potential difference   and i = current through the semiconductor.

     From the given graph, we find that 
              di/dV is +ve for the graph between O and A.
              di/dV is +ve for      "    "            "       A and B.
              di/dV is - ve for     "        "         "       B and C.

      Because the current decreases as voltage increases on the curve B to C.
     So Resistance is negative for the graph between points B and C.