
Two charges q and -2q are separated by a distance d. If the electrical intensity at the site of q is E then the electric field intensity at the site of -2q is:
A. E
B. E/2
C. -2E
D. –E/2
Answer
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Hint: Electric Field is directly proportional to intensity E and charge q, but we must remember that electric field and electric field intensity both are the same. And Electric Field is directly proportional to the product of charge, at first we have to derive the equation for intensity at point charge q, consider the intensity of the charge -2q as some variable, then find the intensity at the site –2q, and after that, we have to compare both the equations for intensity at point charge q and -2q and the final value that we will be getting is the electric field intensity at a charge –2q.
Complete step by step answer:
Electric field intensity is a vector quantity. Electric field intensity at a point is actually a force experienced by unit positive charge placed at that point. Its unit is Newton/coulombs.
The formula for electric field intensity,
$E=\frac{F}{q}$
Here E is the electrical intensity at site q,
$\therefore $ $E=\dfrac{-2q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}$ ……….Eq.1
Let us assume that net electric field at the site of -2q = ${E}'$
$\therefore {E}'=\dfrac{q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}$ ……….Eq.2
On solving Eq.1 and Eq.2 we get,
$\frac{E}{{{E}'}}=\dfrac{\frac{-2q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}}{\dfrac{q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}}$
$\Rightarrow \dfrac{E}{{{E}'}}=-2$
$\Rightarrow {E}'=-\dfrac{E}{2}$ ,
Hence option (D) –E/2 is the correct answer.
Additional Information:
An electric field is a region around a charged object (conductor), within which force is exerted on other charged conductors or objects.
Electric field intensity is a measure of the force exerted by a charged conductor on other charged objects in its field.
Note:
Visualization of charges across two points are required, comparison properties are needed, and a knowledge of Electric field, intensity, charges are required. Students generally consider both the point charges, but in case of an electric field, only the charge that is at the other end should be considered.
Complete step by step answer:
Electric field intensity is a vector quantity. Electric field intensity at a point is actually a force experienced by unit positive charge placed at that point. Its unit is Newton/coulombs.
The formula for electric field intensity,
$E=\frac{F}{q}$
Here E is the electrical intensity at site q,
$\therefore $ $E=\dfrac{-2q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}$ ……….Eq.1
Let us assume that net electric field at the site of -2q = ${E}'$
$\therefore {E}'=\dfrac{q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}$ ……….Eq.2
On solving Eq.1 and Eq.2 we get,
$\frac{E}{{{E}'}}=\dfrac{\frac{-2q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}}{\dfrac{q}{4\pi {{\xi }_{\circ }}{{d}^{2}}}}$
$\Rightarrow \dfrac{E}{{{E}'}}=-2$
$\Rightarrow {E}'=-\dfrac{E}{2}$ ,
Hence option (D) –E/2 is the correct answer.
Additional Information:
An electric field is a region around a charged object (conductor), within which force is exerted on other charged conductors or objects.
Electric field intensity is a measure of the force exerted by a charged conductor on other charged objects in its field.
Note:
Visualization of charges across two points are required, comparison properties are needed, and a knowledge of Electric field, intensity, charges are required. Students generally consider both the point charges, but in case of an electric field, only the charge that is at the other end should be considered.
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