Learn formula of motion of a charged particle in a uniform electric field, Practice calculation of maximum height of charge projected in uniform electric field.
When angle between \(\vec u\) and \(\vec E\) is \(0°\)
\(\left ( \vec u || \vec E \right)\)
When angle between \(\vec u\) and \(\vec E\) is \(180°\)
Note: Force on a negative charge acts in opposite direction of the field.
A \(2\times 10^5\,m/sec\)
B \(2\times 10^3\,m/sec\)
C \(3\times 10^8\,m/sec\)
D \(0\)
When angle between \(\vec u\) and \(\vec E\) is \(0°\)
\(\left ( \vec u || \vec E \right)\)
When angle between \(\vec u\) and \(\vec E\) is \(180°\)
A \(0\)
B \(\dfrac {1}{3}m\)
C \(\dfrac {10}{3}m\)
D \(300 \,m\)
\(\vec a = \dfrac {q\vec E}{m}\) (downwards)
Maximum Height,
\(H_{max}=\dfrac {u^2sin^2\theta}{2g}\)
Maximum Height,
\(H_{max}=\dfrac{u^2sin^2\theta}{2\left(\dfrac{q\vec E}{m} \right)}\)
Conclusion: This problem can be considered as projectile motion under gravity by replacing acceleration due to gravity (g) with acceleration\(\left(\dfrac {q\vec E}{m}\right)\).
Note: Here, acceleration due to gravity is negligible.
A \(2\,m\)
B \(3\,m\)
C \(2.5\,m\)
D \(1.25\,m\)
Case I
Conclusion: Positive charge experiences the force in the direction of electric field and negative charge experiences the force opposite to the direction of electric field.
Case II
Path of positive particle in a uniform electric field \([\;\vec u\;\nparallel\;\vec a\;]\)
Path of negative particle in a uniform electric field \([\;\vec u\;\nparallel\;\vec a\;]\)
Conclusion: Direction of electric field can be determined by interpreting the direction of velocity and acceleration when the path of particle is known.
For this to happen,
\(d= \dfrac {u^2} {2 \left( \dfrac {q\vec E}{m} \right) }\)
Maximum Height = \(\dfrac {u^2} {2g}\)
replacing \(g\rightarrow \dfrac {q\vec E}{m} \)
Maximum Height = \( \dfrac {u^2} {2 \left( \dfrac {q\vec E}{m} \right) }\)
A \(zero\)
B \(10\, V/m\)
C \(10^3 \,V/m\)
D \(10^5\, V/m\)
\(H_{max}=\dfrac {u^2sin^2\theta}{2 \left( \dfrac {q\vec E}{m}\right)}\) \(\left(\because H_{max}=\dfrac {u^2sin^2\theta}{2g}\right)\)
\(d=\dfrac {u^2sin^2\theta}{2 \left( \dfrac {q\vec E}{m}\right)}\)
\(\vec E=\dfrac {mu^2sin^2\theta}{2qd}\)
A \(zero\)
B \(1500\, V/m\)
C \(15\, V/m\)
D \(150\, V/m\)