Learn Electric Field Due to Point Charge with examples. Calculation of Electric Field in 2-D. Force on Charge in an Electric Field and Component of Electric Field on Equatorial Line of Dipole.
Now, if Q is displaced from its position by a distance d, field doesn't propagate instantaneously. It takes finite time to propagate which is given as :
\(t=\dfrac {r}{c}\)
where,
\(c\) is the speed of light.
A Instantaneously
B After time, t = \(\dfrac {r}{c}\)
C Can't be said
D
\(\vec F =\dfrac {k\,Q\,q}{r^2}\hat r\)
\(\vec E = \dfrac {\vec F}{q}=\dfrac {kQ}{r^2}\hat r\)
NOTE: Test charge q also creates an electric field. But charges don't exert forces on themselves. So, q is only measuring field of charge Q.
\(\vec E\,=\,\dfrac{\,\,\vec F}{\lim\limits_{q\to 0} q}\)
2. Other way is to fix charge Q, then q can be of any magnitude.
NOTE: Electric field at any point in space due to charge Q is independent of magnitude of test charge q.
A 5 × 107 N/C
B 5 × 105 N/C
C 2 × 107 N/C
D 1 × 107 N/C
A \(\vec E_{net}=\dfrac {7\;k\;r}{(r^2+a^2)^{3/2}}\hat i\)
B \(\vec E_{net}=\dfrac {14\;k\;r}{(r^2+a^2)^{3/2}}\hat i\)
C \(\vec E_{net}=\dfrac {2\;k\; r}{(r^2+a^2)^{3/2}}\hat i\)
D \(\vec E_{net}=\dfrac {k\;r}{(r^2+a^2)^{3/2}}\;\hat i\)
\(\vec E = \dfrac {kq}{(x^2+y^2)}\;\hat r\)
where,
\(\hat r=\dfrac {x}{\sqrt{x^2+y^2}}\hat i- \dfrac {y}{\sqrt{x^2+y^2}}\hat j\)
\(\vec E = E_x\;\hat i +E_y\;\hat j\)
\(E_x=\dfrac {k\;q\;x}{(\sqrt{x^2+y^2})^3}\)
\(E_y=\dfrac {-k\;q\;y}{(\sqrt{x^2+y^2})^3}\)
A \(36\times10^7 (4\;\hat i -3\;\hat j)\,N/C\), \(1.44 × 10^9\, N/C\)
B \(36\times10^7 (4\;\hat i +3\;\hat j)\, N/C\), \(2.44 × 10^9\, N/C\)
C \(36\times10^7 (4\;\hat i +6\;\hat j)\,N/C\), \(1.44 × 10^{–9} \,N/C\)
D \(36\times10^7 (4\;\hat i -3\;\hat j)\,N/C\), \( 1.44 × 10^8\, N/C\)
\(\vec E = \dfrac {\vec F} {q}\)
A \(2\hat i+3\hat j \;N\)
B \(25\hat i-30\hat j \;N\)
C \(30\hat i+25\hat j \;N\)
D \(25\hat i+30\hat j \;N\)
Force on charge q
?\(\vec F =q\vec E\)
On X-axis
\(T\;sin\theta=F\) ...(i)
On Y-axis
\(T\;cos\theta=mg\) ....(ii)
Dividing (i) by (ii) both equation
\(tan\theta=\dfrac {F}{mg}\) or \(tan\theta=\dfrac {|q\vec E|}{mg}\)
A \(tan^{-1} \left( \dfrac {1}{3} \right)\)
B \(tan^{-1} \left( \dfrac {1}{2} \right)\)
C \(tan^{-1} \left( 2 \right)\)
D \(tan^{-1} \left( 3 \right)\)
A 1.050 × 106 N/C
B 2.157 × 106 N/C
C 1.159 × 106 N/C
D 3.21 × 106 N/C
\(\vec E_{+q}=\dfrac {1}{4\pi\epsilon_0}\times \dfrac {q}{(\sqrt{r^2+\ell^2})^2}\) (B to P)
and \(\vec E_{-q}=\dfrac {1}{4\pi\epsilon_0}\times \dfrac {q}{(\sqrt{r^2+\ell^2})^2}\) (P to A)
Here, \(|\;\vec E_{+q}\;|=|\;\vec E_{-q}\;|=E\)
\(\vec E_{net}=2E\,cos\theta\;(-\hat i)\)
\(\vec E_{net}=\dfrac {2\times1}{4\pi\epsilon_0}\times \dfrac {Q}{({r^2+\ell^2})}\,cos\theta\;(-\hat i)\)
From the triangle
\(cos\theta=\dfrac {\ell}{\sqrt{r^2+\ell^2}}\)
\(\vec E_{net}=\dfrac {2}{4\pi\epsilon_0}\times \dfrac {Q}{({r^2+\ell^2})}\,\times \dfrac {\ell}{\sqrt{r^2+\ell^2}} \;(-\hat i)\)
\(\vec E_{net}=\dfrac {2\;Q\;\ell} {4\pi\epsilon_0\;{(r^2+\ell^2})^{3/2}} \;(-\hat i)\)
A 880 \((\hat i)\)V/m
B 86 × 107 \((-\hat i)\)V/m
C 98 \((\hat i)\) V/m
D 3 × 105 \((\hat j)\)V/m