Practice electric potential due to a point charge and between two charges. Learn electric potential & potential energy definition, examples, formula.
\(V_P=\dfrac {1}{4\pi\epsilon_0}\dfrac {q}{r}\)
NOTE: Put the value of charge with its sign [ (+) positive or (–) negative].
S.I. Unit:
Unit \(\rightarrow\) Nm/C or Volt
case 1 : when charge is positive
case 2 : when charge is negative
A (VA = VB) > (VC = VD) > (VE = VF) and (VP = VQ) > (VR = VS) > (VU = VT)
B (VA = VB) < (VC = VD) < (VE = VF) and (VP = VQ) < (VR = VS) < (VU = VT)
C (VA = VB) < (VC = VD) < (VE = VF) and (VP = VQ) > (VR = VS) > (VU = VT)
D (VA = VB) > (VC = VD) > (VE = VF) and (VP = VQ) < (VR = VS) < (VU = VT)
\(V=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r}\)
So, \(V_P=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r_1}\)
\(V_S=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r_2}\)
\(V_R=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r_3}\)
\(V_{1}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {q}{(a/\sqrt{2})}\)
\(V_{2}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {2q}{(a/\sqrt{2})}\)
\(V_{3}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {(-2q)}{(a/\sqrt{2})}\)
\(V_{4}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {q}{(a/\sqrt{2})}\)
\(V = V_1 + V_2 + V_3 + V_4\)
\(V=\dfrac {2\sqrt{2}\;q}{4\pi\epsilon_0\;a}\)
Conclusion: Total electric potential is the scalar sum of all electric potential due to individual charge.
A \(30\)\(\sqrt{2}\) \(kV\)
B \(45\)\(\sqrt{2}\) \(kV\)
C \(90\)\(\sqrt{2}\) \(kV\)
D \(60\)\(\sqrt{2}\) \(kV\)
Potential of q_{1} at P, \(V_{P_1}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_1}{r_1}\)
and
Potential of q_{2} at P, \(V_{P_2}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_2}{r_2}\)
\(V_P=V_{P_1}+V_{P_2}\)
Electric potential of q at O,
\(V_{q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r}\)
\(V_{-q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-q)}{r}\)
\(V_{2q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {2q}{r}\)
\(V_{-2q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-2q)}{r}\)
\(V = V_q + V_{-q} + V_{2q} + V_{-2q}\)
\(V\) \(=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {+q}{r}+ \dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-q)}{r}+ \dfrac {1}{4\pi\epsilon_0}\;\dfrac {2q}{r}+ \dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-2q)}{r} \)
\(V = 0\)
Conclusion: If sum of all charges placed symmetrically is zero, then total electric potential at center of symmetry will also be zero.
A \(\dfrac {4q}{4\pi\epsilon_0a}\)
B \(\dfrac {3q}{4\pi\epsilon_0a}\)
C \(\dfrac {-q}{4\pi\epsilon_0a}\)
D \(Zero\)
Definition- Change in electrical potential energy of a system is defined as the negative of work done by electrical forces in changing the configuration of the system.
S.I. Unit: Joule
Now, q_{2} is taken to a point C slowly (so that no potential energy changes into kinetic energy), situated at a distance r_{2} from q_{1}.
Electrical force at P,
\(\vec F = \dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_1q_2}{r^2}\) (A to C)
\(dW = \dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_1q_2}{r^2}\;dr\)
[\(\vec F \) and \(d \vec r\) are in same direction]
\(W_{(B\rightarrow C)}=\int\limits_{r_1}^{r_2}\dfrac {1}{4\pi\epsilon_0}\dfrac {q_1q_2}{r^2}dr\)
\(W_{(B\rightarrow C)}=\dfrac {1}{4\pi\epsilon_0}\times q_1q_2 \left [ \dfrac {1}{r_2}-\dfrac {1}{r_1} \right]\)
\(\left [ \because\int\limits_{r_1}^{r_2}\dfrac {1}{r^2} \;dr= \left [ \dfrac {-1}{r} \right]_{r_1}^{r_2}= \left [ -\dfrac {1}{r_2}+\dfrac {1}{r_1} \right] \right]\)
\(\Delta U=U_C-U_B=-W_{B \rightarrow C}\)
\(\Rightarrow\)\(U_C-U_B=-\dfrac {q_1q_2}{4\pi\epsilon_0} \left [ \dfrac {-1}{r_2}+\dfrac {1}{r_1} \right]\)
\(\Rightarrow \)\(U_C-U_B=\dfrac {q_1q_2}{4\pi\epsilon_0} \left [ \dfrac {1}{r_2}-\dfrac {1}{r_1} \right]\)
\(U_{\infty}=0\)
\(\Delta U=U_r-U_{\infty}\)
\(=U_r-U_{\infty}\)
\(=\dfrac {1}{4\pi\epsilon_0}\times q_1q_2 \left[ \dfrac {1}{r}- \dfrac {1}{\infty} \right]\)
\(U_r=\dfrac {q_1q_2}{4\pi\epsilon_0\;r} \)
\(U_r=\dfrac {q_1q_2}{4\pi\epsilon_0r} \)
Note: Value of charges are considered with sign.
\(U=\dfrac {q_1q_2}{4\pi\epsilon_0r}\)
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