Two charges of 6µC and 9µC are 2.5m apart. Where is the electric field along the line joining them equal to zero? show solutions
Answers
Explanation:
Given :-
Object distance (u)= -50cm
Radius (R)= -60cm [•°• Concave mirror]
focal length (f)= R/2= -60/2= (-30)cm
Size of object = 15cm
To Find :-
We have to find the size of image
Solution :-
Using mirror formula
\underline{\boxed{\it\ \dfrac{1}{f}= \dfrac{1}{v}+\dfrac{1}{u}}}
f
1
=
v
1
+
u
1
Now , by putting these values in the formula
\begin{gathered}\longrightarrow\sf\ \dfrac{1}{-30}= \dfrac{1}{-50}+\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-1}{30}+\dfrac{1}{50}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-5+3}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-2}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ v= \cancel{\dfrac{-150}{2}}\\ \\ \\ \longrightarrow\purple{\sf v= (-75)cm}\end{gathered}
⟶
−30
1
=
−50
1
+
v
1
⟶
30
−1
+
50
1
=
v
1
⟶
150
−5+3
=
v
1
⟶
150
−2
=
v
1
⟶ v=
2
−150
⟶v=(−75)cm
Now we have to find the size of image
By using magnification formula
\underline{\boxed{\sf\ m= \dfrac{-v}{u}= \dfrac{h_i}{h_o}}}
m=
u
−v
=
h
o
h
i
\begin{gathered}\longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{\not{-}(-75)}{\not{-}50}\\ \\ \\ \longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{-(75)}{50}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-(\cancel{75}\times 15)}{\cancel{50}}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-3\times 15}{2}\\ \\ \\ \longrightarrow\sf\ \ h_i= \cancel{\dfrac{-45}{2}}\\ \\ \\ \longrightarrow\sf\ \ h_i= (-22.5)\end{gathered}
⟶
15
h
i
=
−50
−(−75)
⟶
15
h
i
=
50
−(75)
⟶ h
i
=
50
−(
75
×15)
⟶ h
i
=
2
−3×15
⟶ h
i
=
2
−45
⟶ h
i
=(−22.5)
\underline{\bigstar{\textsf{\textbf{\ Size\ of\ image = (-22.5)cm}}}}
★ Size of image = (-22.5)cm
thank you