If Á and B denote the sides of a parallelogram and its area is AB/2, find the angle between A and B
Answers
Answer:
the angle between the A and b will be of 1990 degree because they are perpendicular
Answer:
Defination :-
Electric potential at a point may be defined as Work done in bringing a unit positive test charge from infinity to that point without changing kinetic energy.
⇝ In Attached Figure :-
q = Point charge due to which electric potential is to be find.
\large\sf+ q_ \circ+q∘ = +ve Test charge
P = Any point at distance r where potential is to be find.
\large \dag† Derivation :-
[ Figure in Attachment ]
Small work done to move charge from point P to infinity is :
\begin{gathered} \text{dW = - F.dx} \\ \end{gathered}dW = - F.dx
\begin{gathered}:\longmapsto \rm{dW = - q_{\circ}E .dx} \\ \end{gathered}:⟼dW=−q∘E.dx
\begin{gathered}:\longmapsto\text{dW} = - \frac{\text{kqq}_{\circ}}{\text x {}^{2} } \\ \end{gathered}:⟼dW=−x2kqq∘
⏩ Integrating Both Side ;
\begin{gathered}\text{W}_{\text P\rightarrow \infty } = - \int\limits^ \infty _\text r \text{kqq}_{\circ}( {\text x}^{ - 2} ) \ \text {dx}\\ \end{gathered}WP→∞=−r∫∞kqq∘(x−2) dx
\begin{gathered} = - \text{kqq}_{\circ} {{\bigg[ - \frac{1}{\text x} \bigg]}_\text r^ \infty } \\ \end{gathered}=−kqq∘[−x1]r∞
\begin{gathered}\text{W}_{\text P \rightarrow \infty } = - \frac{\text{kqq}_{\circ}}{\text r} \\ \end{gathered}WP→∞=−rkqq∘
\large\red{\therefore \: \boxed{ \boxed{\text{W}_{\infty \rightarrow\text P } = \frac{\text{kqq}_{\circ}}{\text r} }}}∴W∞→P=rkqq∘
⏩ By Defination :
\begin{gathered}:\longmapsto \rm V_P=\frac{ W_{\infty \rightarrow P} }{q_{\circ}} \\\end{gathered}:⟼VP=q∘W∞→P
\begin{gathered}:\longmapsto \rm V_P = \frac{{kqq}_{\circ}}{rq_{\circ}} \\ \end{gathered}:⟼VP=rq∘kqq∘
\purple{ \large :\longmapsto \underline {\boxed{{\bf V_P = \frac{{kq}}{r}} }}}:⟼ VP=rkq
which is electric potential due to a Positive point charge at point P which is r distance away from Positive Charge.