Question

potential energy of a system of two charges in an external field​

Answer 1

Explanation:

Potential energy of a system of two charges in an external field - example. ... Electric field in a region is given as E =2i^. A charge of 1 C is placed at x=1m and a charge of 2 C is placed at x=2 m.

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Answer 2

$\huge{\underline{\rm{Solution:-}}}$

• Let the two charges be q1 and q2 located at r1 and r2 respectively.

Now, work done in bringing q1 at point r1

• → q1V(r1)

The work done in bringing q2 to r2. In this step work is done not only against the external field nut also against the field due to q1:

Workdone on q2 against the external field

• → q2V(r2)

Work done on q2 against the field due to q1

$\longrightarrow \: \tt{\frac{q _{1} \times q _{2} }{4 \pi \epsilon _{0} r _{12} } } \\ \\ \sf{where \: r _{12} \: is \: the \: distance} \\ \sf{between \: q_{1} \: and \: q_{2}}$

By the superposition principle for fields we add the workdone on q2 against the two fields:

Workdone done in bringing q2 to r2:

$\longrightarrow \: \tt{ q _{2} V(r_{2}) + \frac{q _{1} \times q _{2} }{4 \pi \epsilon _{0}r _{ 12} } }$

Thus ,

potential energy of the system= the total work done in assembling the configuration:

$\longrightarrow \: \tt{ q _{1} V(r_{1}) + q _{2} V(r _{2}) + \frac{q _{1} \times q _{2} }{4 \pi \epsilon _{0}r _{ 12} } }$