91影视

• Radius of the disk,$R=2.00\text{鈥塩尘}$
• Charge distributed uniformly on one face of the disk,$Q=+\left(2.00x{10}^6\right)e$
• Width of the circular ring, centered on one face is$w=30\text{鈥坝纠}$
• Radius of the ring,$r=0.50\text{cm}$

Using the concept of surface charge density, the relation on a surface with the density at that surface can be given. This determines the charge contained in the width of the ring.

Formula:

The surface charge density of a distribution, $蟽=dq/dA$ (i)

We know that the area of the circular disk is given as:

$A=蟺r虏$

$鈬�dA=2蟺rdr$鈥︹�� (a)

Now, the charge of an infinitesimal charge$dq$is given using this small area value and equation (i) as:

role="math" localid="1661915197346" $dq=\left(\frac{Q}{蟺R^2}\right)2蟺rdr$

Now, substituting the given values in the above equation for$dr$being the width of the ring, we get the charge in the width as:

$\begin{array}{c}dq=\left(\frac{2脳{10}^6脳1.6脳{10}^{鈭�19}\text{C}}{蟺{\left(0.02m\right)}^2}\right)2蟺\left(0.005\text{鈥尘}\right)(30脳{10}^{鈭�6}\text{鈥尘})\\ =2.4x{10}^{鈭�16}\text{鈥塁}\end{array}$

Hence, the value of the charge in the width is.$2.4x{10}^{鈭�16}\text{鈥塁}$