Due to a charge inside the cube, the electric field is: `E_(x)` = 600x, `E_(y)` = 0, `E_(z)` = 0. The charge inside the cube is nearly - A. 600`mu`CB. 60`mu`CC. 53`mu`CD. 6`mu`C

Due to a charge inside the cube, the electric field is: `E_(x)` = 600x

1.	Due to a charge inside the cube, the electric field is: `E_(x)` = 600x, `E_(y)` = 0, `E_(z)` = 0. The charge inside the cube is nearly - A. 600`mu`CB. 60`mu`CC. 53`mu`CD. 6`mu`C
Answer» Correct Answer - C `q=varepsilon_(0)(Q_(2)-Q_(1))` `phi_(2)=E_(2).A` `phi_(2)=600xx0.2xx0.01=1.2` `phi_(1)=600xx0.1xx0.01=0.6` `q=8.85xx10^(-12)xx0.6` `q=53.10xx10^(-12)`

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Due to a charge inside the cube, the electric field is: `E_(x)` = 600x, `E_(y)` = 0, `E_(z)` = 0. The charge inside the cube is nearly - A. 600`mu`CB. 60`mu`CC. 53`mu`CD. 6`mu`C

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