Consider the charges \(q, q\) and \(-q\) placed at the vertices of an equilateral triangle, as shown in figure. What is the force on each charge?
Speed
00:00
05:25
Consider the charges \(q, q\) and \(-q\) placed at the vertices of an equilateral triangle, as shown in figure. What is the force on each charge?
Answer
By observing the figure
Force acting on charge \(q\) at \(A\) due to charge \(q\) at \(B\) is \(F_{12}\) along \(\mathit{BA}\) as both charges are of same nature therefore repulsive in nature.
Force acting on charge \(q\) at \(A\) due to charge \(-q\) at \(C\) is \(F_{13}\) along \(\mathit{AC}\) as both charges are opposite in nature therefore attracts each other.
By the law of parallelogram, the total force \(F_1\) on the charge \(q\) at \(A\) can be found as,
\(F_1=Fȓ_1\)
where as \(ȓ_1\) is a unit vector along \(\mathit{BC}\) which is at an angle of \(120^{\circ}\) from \(x\)- axis .
The force of attraction or repulsion for each pair of charges \((\)either \(\mathit{AB}\) or \(\mathit{BC}\) or \(\mathit{CA})\) has the same magnitude and therefore,
\(F=\frac 1{4\pi \varepsilon _0}\frac{q^2}{l^2}\)
where the magnitude of charge is \(q\) and the distance between each pair is \(l\) .
The total force \(F_2\) on charge \(q\) at \(B\) is therefore \(F_2=Fȓ_2\) , where \(ȓ_2\) is a unit vector along \(\mathit{AC}\) .
And the total force on charge \(-q\) at \(C\) is \(F_3=\sqrt 3F\widehat n\) , where \(\widehat n\) is the unit vector along the direction bisecting the \({\angle}\mathit{BCA}\) .
Force acting on charge \(q\) at \(A\) due to charge \(q\) at \(B\) is \(F_{12}\) along \(\mathit{BA}\) as both charges are of same nature therefore repulsive in nature.
Force acting on charge \(q\) at \(A\) due to charge \(-q\) at \(C\) is \(F_{13}\) along \(\mathit{AC}\) as both charges are opposite in nature therefore attracts each other.
By the law of parallelogram, the total force \(F_1\) on the charge \(q\) at \(A\) can be found as,
\(F_1=Fȓ_1\)
where as \(ȓ_1\) is a unit vector along \(\mathit{BC}\) which is at an angle of \(120^{\circ}\) from \(x\)- axis .
The force of attraction or repulsion for each pair of charges \((\)either \(\mathit{AB}\) or \(\mathit{BC}\) or \(\mathit{CA})\) has the same magnitude and therefore,
\(F=\frac 1{4\pi \varepsilon _0}\frac{q^2}{l^2}\)
where the magnitude of charge is \(q\) and the distance between each pair is \(l\) .
The total force \(F_2\) on charge \(q\) at \(B\) is therefore \(F_2=Fȓ_2\) , where \(ȓ_2\) is a unit vector along \(\mathit{AC}\) .
And the total force on charge \(-q\) at \(C\) is \(F_3=\sqrt 3F\widehat n\) , where \(\widehat n\) is the unit vector along the direction bisecting the \({\angle}\mathit{BCA}\) .
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
Correct11
Incorrect0
Still Have Question?
Watch More Related Solutions
An electron falls through a distance of \(1.5\;\text{cm}\) in a uniform electric field of magnitude \(2.0\times 10^4NC^{-1}.\) The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance \(.\) Compute the time of fall in each case. Contrast the situation with that of ‘free fall under gravity’.
Two point charges \(q_1\) and \(q_2\) , of magnitude \(+10^{-8}C\) and \(-10^{-8}C\) respectively, are placed \(0.1m\) apart. Calculate the electric fields at points \(A,B\; \)and \(C\) shown in figure.
An electric field is uniform, and in the positive \(x\) direction for positive \(x\) , and uniform with the same magnitude but in the negative \(x\) direction for negative \(x\) . It is given that \(E=200\;\widehat iN/C\)
for \(x>0\) and \(E=\text{-}200\;\widehat iN/C\) for \(x<0\) . A right circular cylinder of length \(20\;\mathrm{cm}\) and radius \(5\;\mathrm{cm}\) has its centre at the origin and its axis along the \(x\) -axis so that one face is at \(x=+10\;\mathrm{cm}\) and the other is at \(x=\text{-}10\;\mathrm{cm}(\text{Figure}).\) What is the net outward flux through the cylinder?
for \(x>0\) and \(E=\text{-}200\;\widehat iN/C\) for \(x<0\) . A right circular cylinder of length \(20\;\mathrm{cm}\) and radius \(5\;\mathrm{cm}\) has its centre at the origin and its axis along the \(x\) -axis so that one face is at \(x=+10\;\mathrm{cm}\) and the other is at \(x=\text{-}10\;\mathrm{cm}(\text{Figure}).\) What is the net outward flux through the cylinder?
An electric field is uniform, and in the positive \(x\) direction for positive \(x\) , and uniform with the same magnitude but in the negative \(x\) direction for negative \(x\) . It is given that \(E=200\widehat iN/C\) for \(x>0\) and \(E=\text{-}200\widehat iN/C\) for \(x<0\) . A right circular cylinder of length \(20\;\text{cm}\) and radius \(5\;\text{cm}\) has its centre at the origin and its axis along the \(x\) -axis so that one face is at \(x=+10\; \text{cm}\) and the other is at \(x=\text{-}10\; \text{cm}\) . What is the net charge inside the cylinder?
An early model for an atom considered it to have a positively charged point nucleus of charge \(\mathit{Ze},\) surrounded by a uniform density of negative charge up to a radius \(R\) . The atom as a whole is neutral. For this model, what is the electric field at a distance \(r\) from the nucleus?
Two sources \(S_{1}\) and \(S_{2}\) emitting light of wavelength \(600\ nm\) are placed a distance \(1.0\times 10^{-2} cm\) apart. A detector can be moved on the line \(S_{1}P\) which is perpendicular to \(S_{1}S_{2}\). What would be the minimum and maximum path difference at the detector as it is moved along the line \(S_1P\)?
Load More
More Solution Recommended For You