Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface, 
Surface charge density of second surface, 
The electric field at a point between the two surfaces is given by :



E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.
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<span>Work=effective force x distance = 300cos36. 100 ft.lb.</span>
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