The electric field vector is zero at point (3). As `-dV//dr = E_r`, the negative of the slope of V verus r curve represents the component of electric field along r . Slope of curve is zero only at `3`.
Near positive charge, net potential is positive and near a negative charge net potential is negative. Thus, charge `Q_1` is positive and `Q_2` negative. From the graph, it can be seen that net potential due to the two charges is positive everywhere in the region left of charge `Q_1`. Therefore, the magnitude of potential due to charge `Q_1` is greater than that due to `Q_2`. Therefore, the absolute value of charge `Q_1` is greater than that of `Q_2`. Secondly, point 1, where potential due to two charges is zero, is nearer to charge `Q_2` thereby implying that `Q_1` has greater absolute value. Also, potential is zero at 2, which is toward right of `Q_2`, as we know that potential is zero at an outside point toward the side of charge smaller in magnitude.