Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of –2 x 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).

Given charge q = 8 mC = 8 x 10–1 C is located at origin and the small charge (q0 = –2 x 10–9 C) is taken from point P (0, 0, 3 cm) to a point Q (0, 4, cm, 0) through point R (0, 6 cm, 9 cm) which is shown in the figure.
Initial separation between q0 and q is rp = 3 cm = 0.03 m
Final separation between q0 and q is rQ = 4 cm = 0.4 m

Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

Work done in taking the charge q0 from point P to Q does not dependent on the path followed and depends only upon rp and rQ i.e., initial and final positions.
                      
Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

or,                  
Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

Two thin dielectric slabs of dielectric constants K1 and K2 K1

Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Two dielectric slab of dielectric constant k1 and k2 and of same thickness is inserted in parallel P

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Suggest Corrections

0