The figure below shows a cross section through a very large nonconducting slab ofthickness d=9.40 mm and uniform volume charge density p=5.80 fC/m3. The originof an x axis is at the slab’s centre. What is the magnitude of the slab’s electric fieldat an x coordinate of (a) 0, (b) 2.00 mm, (c) 4.70mm, (d) 26 mm?
Shubham Rai , 7 Years ago
Grade 12th pass
1 Answers
Vikas TU
Last Activity: 7 Years ago
Volume charge density = 5.80 fC/m3
Thickness d = 9.40 mm
At 0, Electric field will be zero.
At 2 mm
The Electric field will be
E.F. = volume density * volume
= 5.80 * 18.8
=109.04
At 4.70 mm
Electric Field will be
E.F. = volume density* volume
= 5.80* 9.40
= 256.244
At 26mm
E.F. = volume density * volume
= 5.80 * 244.4
= 1417.52
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