A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm). - Physics

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Numerical

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

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Solution

Charge located at the origin, q = 8 mC = 8 × 10−3 C

Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = −2 × 10−9 C

All the points are represented in the given figure.

Point P is at a distance, d1 = 3 cm, from the origin along the z-axis.

Point Q is at a distance, d2 = 4 cm, from the origin along the y-axis.

Potential at point P, `"V"_1 = "q"/(4piin_0 xx "d"_1)`

Potential at point Q,`"V"_2 = "q"/(4piin_0 xx "d"_2)`

Work done (W) by the electrostatic force is independent of the path.

∴ `"W" = "q"_1 ["V"_2 - "V"_1]`

= `"q"_1["q"/(4piin_0"d"_2) - "q"/(4piin_0"d"_1)]`

= `("qq"_1)/(4piin_0) [1/"d"_2 - 1/"d"_1]` .......(1)

Where, `1/(4piin_0) = 9 xx 10^9  "N m"^2  "C"^-2`

∴ `"W" = 9 xx 10^9 xx 8 xx 10^-3 xx (-2 xx 10^-9)[1/0.04 - 1/0.03]`

= `-144 xx 10^-3 xx ((-25)/3)`

= 1.27 J

Therefore, work done during the process is 1.27 J.

Concept: Potential Energy in an External Field - Potential Energy of a Dipole in an External Field
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Chapter 2: Electrostatic Potential and Capacitance - Exercise [Page 87]

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NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.12 | Page 87
NCERT Physics Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 12 | Page 88
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