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capacitor in parallel with contacts 

The direction of the lines of force at any point is the direction along which a unit positive charge placed at that point would move if free to do so. So, the correct option is:

positive

The time constant \( \tau \) of the charging circuit, which represents the time taken for the capacitor to charge to about 63.2% of its maximum voltage when charging through a resistor, is given by:

\[ \tau = RC \]

So, the correct option is:

RC

The dielectric constant \( \varepsilon_r \) is the relative permittivity of a material compared to vacuum or air. It quantifies how much greater the electrostatic field in the material is compared to that in a vacuum or air. 

Given that the electrostatic field in the dielectric is 50 times greater than in air, we can use the formula:

\[ \varepsilon_r = \dfrac{{E_{\text{{dielectric}}}}}{{E_{\text{{air}}}}} \]

where \( E_{\text{{dielectric}}} \) is the electrostatic field in the dielectric and \( E_{\text{{air}}} \) is the electrostatic field in air.

Since \( \varepsilon_r = \dfrac{{E_{\text{{dielectric}}}}}{{E_{\text{{air}}}}} = 50 \), the dielectric constant of the dielectric will be 50.

So, the correct option is:

50

For a good 0.05 μF capacitor, an ohmmeter reading should show high resistance momentarily and then a very low resistance. This behavior occurs because the capacitor initially looks like an open circuit (high resistance) until it charges, after which it looks like a short circuit (very low resistance) to the DC current supplied by the ohmmeter.

In the case of a lossy capacitor, its series equivalent resistance value will be relatively large. Lossy capacitors have inherent resistance due to factors such as dielectric losses and internal conductor resistance. This resistance causes energy dissipation in the form of heat when the capacitor is charged and discharged.

So, the correct option is:

large

If the dielectric of a capacitor is replaced by a conducting material, the plates of the capacitor will effectively get short-circuited. This is because the conducting material will provide a low resistance path between the plates, allowing charge to flow freely. As a result, the capacitance value of the capacitor will decrease significantly, and the plates will essentially lose their ability to store charge. 

So, the correct option is:

the plates will get short-circuited

When an uncharged conductor is placed near a charged conductor, it undergoes a process called electrostatic induction. In this process, the charges in the uncharged conductor redistribute themselves under the influence of the electric field produced by the charged conductor. The redistribution of charges creates an induced opposite charge on the side of the uncharged conductor facing the charged conductor and an induced like charge on the opposite side. 

So, the correct option is:

the uncharged conductor gets charged by induction and then attracted towards the charging body

The time constant \( \tau \) of an RC circuit (resistor-capacitor circuit) is given by the product of the resistance \( R \) and the capacitance \( C \), i.e., \( \tau = RC \). 

Given:
- Capacitance \( C = 100 \mu F \)
- Resistance \( R = 8000 \Omega \)

So, the time constant \( \tau \) is:\[ \tau = RC = (100 \times 10^{-6}) \times 8000 \]\[ \tau = 0.8 \, \text{s} \]

Thus, the time constant of the circuit is 0.8 seconds. None of the provided options match this value. It seems there might be an error in the options provided.

The phenomenon of an uncharged body getting charged merely by the nearness of a charged body is known as electrostatic induction. So, the correct term is:

Induction