Question

What is the force experienced by a test charge of \[0.20\,\mu C\] placed in an electric field of \[3.2 \times {10^6}\,N{C^{ - 1}}\].

Answer 1

Hint: To answer this question we will first know about what electrostatic force of attraction or repulsion is in a case where a point charge is placed in an electric field.Later we will apply a simple formula that relates electrostatic force, charge and electric field.

Formula used:
\[F = QE\]
Where \[F\] = electrostatic force , \[Q\]= charge and \[E\]= electric field.

Complete step by step answer:
Let us look at what electrostatic force of attraction or repulsion is ,in a case where a point charge is placed in an electric field. Electric force is the pull or push that an electric charge will experience when placed in an electric field. There is another law which gives information about electric force in an electric field called coulomb’s law, but here we will not apply that as all terms needed in coulomb’s law formula are not mentioned.

An electric force is the repulsive or attractive interaction between any two charged particles.
Now lets see what all has been given to us:
\[Q = 0.20\,\mu C\]
\[ \Rightarrow Q = \left( {0.20 \times {{10}^{ - 6}}} \right)C\]
\[E = 3.2 \times {10^6}N{C^{ - 1}}\]
The force of attraction or repulsion in case of a point charge placed in an electric field is given by a simple formula :
\[F = QE\]
\[ \Rightarrow F = \left( {0.20 \times {{10}^{ - 6}}} \right) \times \left( {3.2 \times {{10}^6}} \right)\]
\[ \Rightarrow F = 0.20 \times 3.2\,N\]
\[ \therefore F = 0.64\,N\]

Hence the force experienced by a test charge of \[0.20\,\mu C\] placed in an electric field of \[3.2 \times {10^6}\,N{C^{ - 1}}\] is \[0.64\,N\].

Note:Remember to convert the unit of charge from microcoulomb to coulomb to answer this question in N/C. The force that we calculate can be both attractive or repulsive in nature. But in this question the force given is positive hence we know that the force is attractive in nature.